Introduction to Thermal and Fluid Engineering 9781466503212, 1466503211

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2011 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20120111 International Standard Book Number-13: 978-1-4665-0321-2 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Dedicated to Fred Landis, a fine friend, a valued colleague, and a superb servant of our profession for which he never received the credit that he deserved.

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Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xix

1. The Thermal/Fluid Sciences: Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5 Engineered Systems and Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.1 Case I: A Commercial Airliner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.2 Case II: A Chemical Plant Distillation Column . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5.3 Case III: The Thermal Ink Jet Printer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.6 Historical Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6.1 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6.2 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6.3 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.7 The Thermal/Fluid Sciences and the Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.7.1 The Ozone Layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.7.2 Smog. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 1.7.3 Acid Rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2. Thermodynamics: Preliminary Concepts and Definitions. . . . . . . . . . . . . . . . . . . . . . . . .15 2.1 The Study of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Some Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.2.1 Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.2.2 Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2.3 Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2.4 State and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.2.5 Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.2.6 Thermodynamic Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2.6.1 Requirements for Thermodynamic Equilibrium . . . . . . . . . . . . . 19 2.2.6.2 Quasi-Equilibrium or Quasi-Static Processes . . . . . . . . . . . . . . . . 20 2.2.7 Statistical and Classical Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.3 Dimensions and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3.2 The SI (Syst e` me Internationale D’unit e´ s) System . . . . . . . . . . . . . . . . . . . . . 21 2.3.3 The English Engineering System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.3.3.1 The British Gravitational System . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3.4 SI Unit Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.4 Density and Related Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.5 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

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2.6 2.7 2.8 2.9

2.5.1 Gage, Absolute, and Vacuum Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.5.2 Measuring Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Temperature and the Zeroth Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 29 Problem-Solving Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3. Energy and the First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2 Kinetic, Potential, and Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.2.1 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.2.2 Potential Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39 3.2.3 Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.2.4 The Total Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.3 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 3.3.1 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.4 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.5 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.6 The Energy Balance for Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.6.1 Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.6.2 Other Forms of Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.6.2.1 Shaft or Paddle Wheel Work. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .50 3.6.2.2 Electrical Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.6.2.3 Spring Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.6.2.4 Two Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 3.6.3 Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.7 The Ideal Gas Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.8 Ideal Gas Enthalpy and Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.9 Processes of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.9.2 The Constant Volume Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.9.3 The Constant Pressure Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.9.4 The Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 3.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 4. Properties of Pure, Simple Compressible Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.1 The State Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.2 P-v-T Relationships. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74 4.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.2.2 The P-v-T Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.2.3 Single-Phase Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.2.4 Two-Phase Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 4.2.5 Three-Phase Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 4.2.6 The P-T Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 4.2.7 The P-v Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 4.2.8 Some Hints For Phase Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.3 Thermodynamic Property Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.3.1 Table Arrangement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.3.2 The Saturation Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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4.4 4.5

4.6

4.7

4.8 4.9

ix 4.3.3 The Superheated Vapor Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 4.3.4 The Compressed or Subcooled Liquid Region . . . . . . . . . . . . . . . . . . . . . . . . 86 The T-s and h-s Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Real Gas Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.5.2 The Compressibility Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 4.5.3 The Principle of Corresponding States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 4.6.1 The Van der Waals Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 4.6.2 The Redlich-Kwong Equation of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 The Polytropic Process for an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 4.7.1 P-V-T Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 4.7.2 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 4.7.3 Internal Energy and Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 4.7.4 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5. Control Volume Mass and Energy Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 5.2 The Control Volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110 5.3 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 5.3.1 The Mass Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 5.3.2 The Volumetric Flow Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 5.4 Conservation of Energy for a Control Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 5.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 5.4.2 The Energy Rate Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 5.4.3 Flow Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 5.4.4 The Control Volume Energy Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .117 5.5 Specific Heats of Incompressible Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 5.6 Applications of Control Volume Energy Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 5.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 5.6.2 The Nozzle and the Diffuser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 5.6.3 The Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 5.6.4 The Compressor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .124 5.6.5 Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 5.6.6 The Mixing Chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 5.6.7 Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 5.6.8 The Throttling Valve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 5.7 Synthesis or Analysis? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 5.8 The First Law Heat Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 5.9 Design Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 5.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 5.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 6. The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 6.2 The Kelvin-Planck Statement and Heat Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 6.3 The Clausius Statement: Refrigerators and Heat Pumps . . . . . . . . . . . . . . . . . . . . 154 6.4 The Equivalence of the Kelvin-Planck and Clausius Statements . . . . . . . . . . . . . 158

x

Contents 6.5 6.6 6.7 6.8 6.9 6.10

Reversible and Irreversible Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 The Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 The Carnot Cycle with External Irreversibilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 The Absolute Temperature Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

7. Entropy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .183 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 7.2 The Classical Definition of Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .184 7.3 The Clausius Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 7.4 The Temperature-Entropy Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 7.5 The Gibbs Property Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 7.6 Entropy Change for Solids, Liquids, and Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . 192 7.6.1 Entropy Change for Solids and Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.6.2 Entropy Change for an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 7.7 The Isentropic Process for an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 7.8 Isentropic Efficiencies of Steady Flow Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 7.9 The Entropy Balance Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 7.9.1 The Entropy Balance for Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 7.9.2 The Entropy Rate Balance for an Open System (Control Volume) . . . . 207 7.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 7.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 8. Gas Power Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .219 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 8.2 The Internal Combustion Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 8.3 The Air Standard Otto Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 8.3.1 Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 8.3.2 The Compression Ratio and Its Effect on Performance . . . . . . . . . . . . . . 229 8.4 Design Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 8.5 The Air Standard Diesel Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 8.6 The Gas Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 8.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 8.6.2 The Ideal Gas Turbine or Brayton Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 8.6.3 The Ideal Brayton Cycle with Regeneration . . . . . . . . . . . . . . . . . . . . . . . . . 243 8.7 The Jet Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 8.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 8.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 9. Vapor Power and Refrigeration Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 9.2 The Steam Power Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 9.3 The Ideal Rankine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 9.3.1 The Ideal Rankine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 9.3.2 Increasing the Efficiency of the Ideal Rankine Cycle . . . . . . . . . . . . . . . . . 268 9.4 The Ideal Rankine Cycle with Superheat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 9.5 The Effect of Irreversibilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 9.6 The Rankine Cycle with Superheat and Reheat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 9.7 Design Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 9.8 The Ideal Rankine Cycle with Regeneration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

Contents 9.9

9.10 9.11 9.12 9.13

xi The Ideal Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 9.9.1 Physical Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .286 9.9.2 Refrigerants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 The Ideal Vapor Compression Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . . . . . . 288 Departures from the Ideal Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

10. Mixtures of Gases, Vapors, and Combustion Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 10.2 Mixtures of Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 10.2.1 Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 10.2.2 Dalton’s Law of Partial Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 10.2.3 Gravimetric and Volumetric Analyses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 10.2.4 The Apparent Molecular Weight and Gas Constant . . . . . . . . . . . . . . . . . 305 10.2.5 Properties of Ideal Gas Mixtures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .306 10.3 Psychrometrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 10.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 10.3.2 Specific and Relative Humidity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .309 10.3.3 Other Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 10.3.4 The Wet Bulb Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 10.4 The Psychrometric Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 10.5 The Products of Combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 10.5.1 Fuels. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .316 10.5.2 The Combustion Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 10.5.3 Combustion Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 10.5.4 The Air-Fuel Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 10.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 10.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 11. Introduction to Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 11.1 The Definition of a Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 11.2 Fluid Properties and Flow Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 11.3 The Variation of Properties in a Fluid. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .331 11.4 The Continuum Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 11.5 Laminar and Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 11.6 Fluid Stress Conventions and Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 11.7 Viscosity, a Fluid Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 11.8 Design Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 11.9 Other Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 11.9.1 Specific Gravity and Specific Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 11.9.2 Compressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 11.9.3 The Speed of Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 11.9.4 Surface Tension and Capillary Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 11.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 11.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 12. Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 12.2 Pressure Variation in a Static Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 12.3 Hydrostatic Pressure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .363

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Hydrostatic Forces on Plane Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 Design Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 Hydrostatic Forces on Curved Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 Uniform Rectilinear Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

13. Control Volume Analysis—Mass and Energy Conservation . . . . . . . . . . . . . . . . . . . . . 407 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 13.2 Fundamental Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 13.3 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 13.4 Mass Conservation Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 13.5 The First Law of Thermodynamics for a Control Volume . . . . . . . . . . . . . . . . . . . 414 13.6 Applications of the Control Volume Expression for the First Law . . . . . . . . . . . 415 13.7 The Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 13.8 Design Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 13.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 13.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 14. Newton’s Second Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 14.2 Linear Momentum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .441 14.3 Applications of the Control Volume Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 14.4 Design Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 14.5 The Control Volume Relation for the Moment of Momentum . . . . . . . . . . . . . . . 454 14.6 Applications of the Moment of Momentum Relationship . . . . . . . . . . . . . . . . . . . 456 14.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460 14.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460 15. Dimensional Analysis and Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 15.2 Fundamental Dimensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .477 15.3 The Buckingham Pi Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 15.4 Reduction of Differential Equations to a Dimensionless Form . . . . . . . . . . . . . . . 483 15.5 Dimensional Analysis of Rotating Machines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .485 15.6 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 15.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492 15.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492 16. Viscous Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 16.2 Reynolds’ Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 16.3 Fluid Drag. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .505 16.4 Design Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 16.5 Boundary Layer Flow over a Flat Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 16.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516 16.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517

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17. Flow in Pipes and Pipe Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523 17.2 Frictional Loss in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 17.3 Dimensional Analysis of Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 17.4 Fully Developed Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527 17.5 Friction Factors for Fully Developed Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .528 17.5.1 Laminar Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 17.5.2 Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 17.6 Friction Factor and Head Loss Determination for Pipe Flow . . . . . . . . . . . . . . . . 530 17.6.1 Pipe Friction Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530 17.6.2 Head Loss Due to Fittings and Valves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530 17.6.3 Noncircular Flow Passages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534 17.6.4 Single-Path Pipe Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534 17.7 Design Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541 17.8 Design Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 17.9 Design Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 17.10 Multiple-Path Pipe Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 17.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 17.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 18. Fluid Machinery. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .563 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563 18.2 The Centrifugal Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 18.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 18.2.2 Theoretical Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 18.3 The Net Positive Suction Head . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 18.4 Combining Pump and System Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 18.5 Scaling Laws for Pumps and Fans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576 18.6 Axial and Mixed Flow Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 18.7 Turbines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 18.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 18.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582 19. Introduction to Heat Transfer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .591 19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 19.2 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 19.3 Thermal Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 19.4 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 19.5 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 19.6 Thermal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 19.7 Combined Mechanisms of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598 19.8 The Overall Heat Transfer Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601 19.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 19.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 20. Steady-State Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609 20.2 The General Equation of Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610

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20.4

20.5

20.6

20.7

20.8 20.9

Conduction in Plane Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613 20.3.1 The Single-Material Layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613 20.3.2 The Composite Plane Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 618 20.3.3 Contact Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 622 Radial Heat Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626 20.4.1 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626 20.4.1.1 The Hollow Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626 20.4.1.2 The Composite Hollow Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . 628 20.4.1.3 The Critical Radius of Insulation . . . . . . . . . . . . . . . . . . . . . . . . . . 631 20.4.2 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 20.4.2.1 The Hollow Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633 20.4.2.2 The Composite Hollow Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634 Simple Shapes with Heat Generation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .636 20.5.1 The Plane Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636 20.5.2 The Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 20.5.3 The Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 640 Extended Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642 20.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642 20.6.2 The Longitudinal Fin of Uniform Thickness . . . . . . . . . . . . . . . . . . . . . . . . 642 20.6.2.1 Constant Base Temperature with Tip Heat Loss. . . . . . . . . . . .644 20.6.2.2 Constant Base Temperature with Insulated Tip . . . . . . . . . . . . 645 20.6.3 Fin Performance Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645 20.6.4 The Cylindrical Spine or Pin Fin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647 20.6.5 Annular or Radial Fin of Uniform Thickness . . . . . . . . . . . . . . . . . . . . . . . . 649 Two-Dimensional Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651 20.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651 20.7.2 Solution Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652 20.7.3 The Conduction Shape Factor Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657

21. Unsteady-State Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 21.2 The Lumped Capacitance Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 21.2.1 Convective Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 21.2.2 The Validity Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669 21.2.3 The Effect of Internal Heat Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671 21.3 The Semi-Infinite Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673 21.3.1 Constant Surface Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674 21.3.2 Constant Surface Heat Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675 21.3.3 Surface Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675 21.4 Design Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677 21.5 Finite-Sized Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 21.5.1 The Long Plane Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 21.5.2 One-Term Approximate Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684 21.5.3 The Long Solid Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686 21.5.4 One-Term Approximate Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691 21.5.5 The Solid Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693 21.5.6 One-Term Approximate Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695 21.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699 21.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 701

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22. Forced Convection—Internal Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 22.2 Temperature Distributions with Internal Forced Convection . . . . . . . . . . . . . . . . 708 22.2.1 The Constant Wall Heat Flux Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708 22.2.2 The Constant Wall Temperature Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711 22.3 Convective Heat Transfer Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714 22.3.1 Case 1—Laminar Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717 22.3.2 Case 2—Transition Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718 22.3.3 Case 3—Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718 22.4 Applications of Internal Flow Forced Convection Correlations . . . . . . . . . . . . . . 718 22.5 Design Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726 22.6 Design Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 730 22.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735 22.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736 23. Forced Convection—External Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745 23.2 Flow Parallel to a Plane Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746 23.2.1 Laminar Boundary Layer Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746 23.2.2 Turbulent Boundary Layer Flow. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .749 23.2.3 Additional Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 750 23.2.3.1 Constant Heat Flux Wall Condition . . . . . . . . . . . . . . . . . . . . . . . 750 23.2.3.2 Unheated Starting Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751 23.3 External Flow over Bluff Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752 23.3.1 The Cylinder in Cross Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752 23.3.2 Tube Bundles in Cross Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 756 23.3.3 Single Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761 23.3.4 Bodies with Noncircular Cross Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761 23.4 Design Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 762 23.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765 23.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766 24. Free or Natural Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773 24.2 Governing Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774 24.3 Working Correlations for Natural Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777 24.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777 24.3.2 Plane Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777 24.3.2.1 Vertical Plates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .777 24.3.2.2 Inclined Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779 24.3.3 Vertical Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779 24.3.4 Horizontal Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781 24.3.5 Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784 24.3.6 Horizontal Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784 24.4 Natural Convection in Parallel Plate Channels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786 24.4.1 The Elenbaas Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786 24.4.2 A Composite Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 788 24.4.3 Optimum Plate Spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790 24.5 Design Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791 24.6 Natural Convection in Enclosures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795 24.6.1 Working Correlations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796

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Contents 24.6.1.1 Vertical Rectangular Enclosures . . . . . . . . . . . . . . . . . . . . . . . . . . . 796 24.6.1.2 Tilted Vertical Enclosures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796 24.6.2 Concentric Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 797 24.6.3 Concentric Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 798 24.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 799 24.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 799

25. Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 25.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 25.2 Governing Relationships. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .807 25.2.1 The Rate Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 807 25.2.2 The Exchanger Surface Area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .807 25.2.3 The Overall Heat Transfer Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808 25.2.4 The Logarithmic Mean Temperature Difference . . . . . . . . . . . . . . . . . . . . . 811 25.2.5 Fouling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815 25.2.5.1 Fouling Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815 25.2.5.2 Fouling Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816 25.3 Heat Exchanger Analysis Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818 25.3.1 The Logarithmic Mean Temperature Difference Correction Factor Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818 25.3.2 The Effectiveness Ntu Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824 25.3.2.1 Dimensionless Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824 25.3.2.2 Specific Effectiveness -Ntu Relationships. . . . . . . . . . . . . . . . . .825 25.4 Design Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 829 25.5 Finned Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834 25.5.1 The Surface Area and the Overall Surface Efficiency . . . . . . . . . . . . . . . . 835 25.5.2 The Overall Heat Transfer Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836 25.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 840 25.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 841 26. Radiation Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 26.1 The Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 26.2 Monochromatic Emissive Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 850 26.2.1 The Black Surface or Blackbody . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 850 26.2.2 Planck’s Law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .851 26.2.3 Wien’s Displacement Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851 26.2.4 The Stefan-Boltzmann Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852 26.3 Radiation Properties and Kirchhoff’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855 26.3.1 Absorptivity, Reflectivity, and Transmissivity . . . . . . . . . . . . . . . . . . . . . . . 855 26.3.2 Kirchhoff’s Radiation Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 856 26.3.3 Emissivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 857 26.3.4 An Approximation to a Blackbody . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858 26.4 Radiation Intensity and Lambert’s Cosine Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 859 26.5 Monochromatic and Total Emissivity and Absorptivity . . . . . . . . . . . . . . . . . . . . . 861 26.5.1 Emissivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861 26.5.2 Absorptivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862 26.5.3 The Gray Surface or Gray Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 26.6 Heat Flow between Blackbodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 26.6.1 The Shape Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 26.6.2 A Catalog of Simple Shape Factors in Two Dimensions . . . . . . . . . . . . . 867 26.6.3 A Catalog of Simple Shape Factors in Three Dimensions . . . . . . . . . . . . 871

Contents

xvii Properties of the Shape Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 26.6.4.1 The Reciprocity Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 26.6.4.2 The Additivity Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 26.6.4.3 The Enclosure Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 26.6.5 The Symmetry Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876 Heat Flow by Radiation between Two Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876 26.7.1 Diffuse Blackbodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876 26.7.2 Opaque Gray Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876 Radiosity and Irradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879 Radiation within Enclosures by a Network Method . . . . . . . . . . . . . . . . . . . . . . . . 880 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885 26.6.4

26.7

26.8 26.9 26.10 26.11

Appendix A: Tables and Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 899 Appendix B: Summary of Differential Vector Operations in Three Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935 References and Additional Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 943 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 947

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Preface

This text treats the disciplines of thermodynamics, fluid mechanics, and heat transfer, in that order, as comprising what are generally referred to as the thermal/fluid sciences. The study of these separate and independent disciplines has been a standard part of the mechanical and chemical engineering curricula for decades. Other engineering majors have commonly taken one or more of these subjects but, generally, not all three. The first component, classical thermodynamics, involves the interaction of work, heat, and the change in the energy level of a system as it undergoes a change between equilibrium states. The laws of thermodynamics form a framework by which these state changes are evaluated and related to measurable properties, most notably temperature. Knowledge of the thermodynamic limits of processes is essential to the evaluation of energetic systems and all engineers need such knowledge upon occasion. The second component, fluid mechanics, treats the change of mass, energy, and momentum associated with the movement of fluids. Mass flow into and out of an open system is an important part of the overall energy balance for the system. Large, multicomponent systems such as municipal water supplies, petroleum refineries, and manufacturing facilities involve numerous pipes, ducts, and other passageways through which fluids are transported and fluid flow analyses are important for describing the rates at which energetic processes take place. The third component of the thermal/fluid sciences is heat transfer. Heat transfer, as one knows from a study of introductory physics is accomplished by conduction, convection, and radiation. Each of these modes enables one to evaluate the rate at which heat is transported between sites that are at different temperatures. Convection heat transfer is intimately involved with fluid motion and is, therefore, directly coupled to a knowledge of fluid mechanics. As mentioned, thermodynamic analysis will determine the limiting equilibrium states that a process may experience. The project design of an energetic process requires, in addition to a listing of the appropriate thermodynamic limits, knowledge of the rates at which the process progresses from its initial to its final states. For example, in the case of a heat exchanger, the cross-sectional area of the unit is evaluated by the rate of mass throughput as determined by using fluid mechanics while the length of the unit is determined from the rate of heat transfer. The foregoing comments speak directly to the interdependence of the three disciplines that comprise the thermal/fluid sciences. This text introduces these subjects to students who, most likely, will not take further course work in this area. We anticipate that the text will be used in a rather intensive one-semester course most likely in the junior year of an engineering curriculum. Knowledge of basic physics and mathematics through ordinary differential equations is assumed. Each chapter includes several example problems, worked in detail, that illustrate the application of the material being introduced. In addition, numerous problem sets are included at the end of each chapter. A uniform, nonintegrated, treatment of the subject matter, using consistent units and nomenclature is one of the features of this text. The authors believe that a fundamental xix

xx

Preface

knowledge of energetic processes and the ability to quantify them is essential for all engineering graduates. We hope that the text will induce more students to acquire these important capabilities in their education and subsequent careers. The means to this end is motivation. The authors have managed to garner diplomas in mechanical, chemical, and electrical engineering, as well as applied mathematics. All have been involved in teaching this material on the undergraduate and graduate levels as well as the “survey course” that you may be taking or are about to take. With regard to this survey course, we have found a rather unusual lack of motivation among students, probably because of the lack of interest in the subject matter, resentment at being required to take such a course, and/or the pressure of other course work that is time consuming and considered more important. As we wrote this book, we anticipated a lack of motivation on the part of the student/reader and have written the book with the student in mind. Accordingly, you may find that the material covered is extremely fascinating. Indeed, by the end of Chapter 5, you will be extremely adroit at the bookkeeping required to keep track of the forms of energy necessary to solve detailed problems involving closed and open thermodynamic systems. We do not lose sight of the difference between system synthesis and system analysis and close to a score of “design problems” have been included. These design problems range from a simple first law “heat balance” to a detailed design of a heat sink to be employed in the cooling of a package of electronics. You will learn how automobile and aircraft engines work, how steam power plants and refrigeration systems are put together, and you will become well versed in such diverse topics as fluid statics, buoyancy, stability, the flow of fluids in pipes and fluid machinery. Finally, the thermal control of electronic components pervade the entire heat transfer section of the book. The authors have tried to anticipate problems that you will encounter in your learning process and this has been their motivation. It is a book that has not been easy to produce. Your motivation is to give the subject matter contained here half a chance and to recognize that while you are sitting in the classroom, after paying a substantial tuition, if you listen, you might just learn something. We have not provided a disk of thermal, fluid, and thermodynamic properties because all of the properties that you will need may be found in the 22 tabulations in Appendix A. Neither have we provided a disk containing any computer programs as we see little point in providing computer codes for a small portion of the problems in the text. The disk that is attached is a comprehensive study guide and is approximately 600 pages. It is intended to help you, to motivate you, and to provide you with ready reference to the material contained herein. We acknowledge the assistance of Jonathan Plant, our editor and an old friend B. J. Clark, who is our editorial consultant. Allan Kraus James Welty Abdul Aziz September 2010

1 The Thermal/Fluid Sciences: Introductory Concepts

Chapter Objectives To introduce the component disciplines that comprise the thermal/fluid sciences—thermodynamics, fluid mechanics, and heat transfer. • To discuss the physical laws upon which the thermal sciences are based. •

•

1.1

To present a list of products and/or processes whose analysis and design rely upon thermal/fluid science principles.

Introduction

The subject areas of thermodynamics, fluid mechanics, and heat transfer comprise what are generally referred to as the thermal/fluid sciences. These subjects are often studied separately but are sufficiently interrelated that they are frequently taken sequentially by engineering students and, as with this text, are often treated in a single book. In this book, these subjects will be treated in a relatively fundamental fashion. Our goal is to provide a basic understanding of the physical laws and processes that involve energy utilization in ways that accomplish useful tasks. Thermal/fluid systems involve energy that can be stored, transferred, or converted within the system. Energy storage manifests itself in different forms such as gravitational or potential and kinetic energy as well as energy that is contained in the matter that constitutes the system. Energy can be transferred between a system and its surroundings via the flow of heat, work, and the flow of hot and/or cold fluid streams of matter. It may be converted from one form to another such as the conversion of the energy contained within fuels to electrical energy as in a power plant, or to mechanical energy, which is used to propel your automobile. A steam power plant (Figure 1.1) converts the chemical energy contained in a fossil fuel to electrical power, while the automobile and the turbojet engines contained in a fuel are converted to propulsive power. The familiar household refrigerator provides a conditioned low temperature environment by employing electrical energy to drive a heat transfer process that involves the use of a refrigerant. Additional applications include the pressure cooker, the human cardiovascular system (Figure 1.2), and the solar panel (Figure 1.3). A study of thermodynamics (in Chapter 4) tells us that the boiling point of water decreases as the surrounding pressure decreases. The pressure cooker affords an opportunity to raise the cooking temperature by means of a regulated pressure. The cardiovascular system is a rather complex combination of 1

2

Introduction to Thermal and Fluid Engineering

Superheater Flue Gas Exhaust

Boiler Flue Gas Out Steam Generator

Generator

Steam

Electric Power Turbine Fuel In Condensate

Pump

Cooling Water In

Cooling Water Out

Condenser

FIGURE 1.1 Artist’s rendering of a steam power plant.

heat transfer and fluid flow components that regulate the flow of air and blood within the relatively narrow range necessary to maintain life. In the solar collector pump and storage system displayed in Figure 1.3, the solar panel provides an inexpensive supply of hot water. Thermal/fluid systems are all around us. Some occur naturally while others are manmade. All are amenable to analysis using a small number of physical laws. These same laws provide the basic building blocks for engineering design. To Body Aorta Right Atrium

Pulmonary Artery To Lungs

Oxygenated Blood from Lungs

Left Atrium

Venous Blood

Right Ventricle

FIGURE 1.2 Artist’s rendering of the human cardiovascular system.

Left Ventricle

3

Ra So d i l ar at io n

The Thermal/Fluid Sciences: Introductory Concepts

Hot Water

Solar Collector

Cold Water Circulating Pump FIGURE 1.3 Artist’s rendering of a solar collector system.

In the remaining portion of this introductory chapter, we will attempt to set the stage by discussing the disciplines of thermodynamics, fluid mechanics, and heat transfer in a general way. More formal statements of these subjects and the laws that they involve will be treated in subsequent chapters. Several examples of engineering systems and manufactured products will be cited. We hope to make it clear that the thermal/fluid sciences comprise an important part of all engineers’ understanding and are essential components for analyzing and/or designing processes that utilize energy to enhance our lives.

1.2

Thermodynamics

Thermodynamics can be generally defined as the study of the relationships between heat, work, and the properties of systems. System properties are those that define energy content. An important feature of thermodynamic analysis is that it relates heat transfer and work done on or by a system as it experiences changes between equilibrium states. The building blocks of thermodynamic analysis include the following: •

The zeroth law of thermodynamics is, essentially, a necessary concept for defining the property we know as temperature.

•

The first law of thermodynamics is a statement that energy must be conserved. This precept is one of the most powerful tools available for the analysis and design of products, processes, and systems.

•

The second law of thermodynamics, which pertains to energy transport being restricted such that processes, while still in compliance with the first law, must obey certain constraints. The system property, entropy, is useful in second law analysis.

To illustrate the concepts involved in the foregoing statements, consider a thermodynamic system consisting of a cup of hot coffee sitting on a kitchen counter as indicated in Figure 1.4. As we know, the heat transfer between the coffee and room air will be directly related to the magnitude of its temperature change. The first law of thermodynamics requires that, in the absence of any work being done on or by the coffee in the cup, the heat exchange

4

Introduction to Thermal and Fluid Engineering

FIGURE 1.4 A cup of coffee resting on a kitchen counter.

with the environment must equal the amount by which the coffee and the cup change in energy content. The first law requirement could be satisfied by either the coffee becoming hotter or cooler. The second law of thermodynamics does not allow heat to transfer from a cooler to a warmer temperatures of its own accord. Hence, the coffee must decrease in temperature.

1.3

Fluid Mechanics

Fluid mechanics is defined as the branch of applied mechanics that involves fluid behavior, either at rest or in motion. All fluid mechanics analyses are based on one or more of the following laws: •

The first law of thermodynamics involves the conservation of energy. Energy conservation is a major consideration when evaluating the effects of fluid flow. Most often, we work with open systems which, by definition, have mass entering and leaving via fluid flow.

•

The conservation of mass is a straightforward and powerful tool and is an absolute requirement for all engineering analysis. Mass conservation equations are often referred to as continuity expressions.

•

Newton’s second law of motion relates force to the time rate of change of momentum. The term momentum equation is often used to refer to a statement of Newton’s second law.

•

The second law of thermodynamics has important consequences when flow is compressible, involving gases and vapors under certain conditions. When flow may be considered to be incompressible, the second law is not a necessary consideration.

Most fluid properties are temperature dependent. For compressible fluids such as gases, pressure is also an important factor. A less familiar property, which is important for describing fluid flow, is the viscosity.

The Thermal/Fluid Sciences: Introductory Concepts

1.4

5

Heat Transfer

We consider heat transfer to be the rate of energy exchange between systems or objects as a result of a temperature difference. Heat transfer occurs in three ways: •

Conduction heat transfer occurs as higher-energy-level molecules (as indicated by higher temperature) transfer thermal energy to lower-energy-level molecules (clearly at lower temperature). Conduction can occur in the steady state (independent of time) or it can be transient (time dependent). Conduction occurs through direct contact between high and low temperature materials.

•

Convection heat transfer is associated with energy exchanged between a surface and an adjacent fluid. The distinction is made between forced and free or natural convection. Forced convection occurs when the fluid flow past the surface is produced by an external agent such as a fan, pump, or blower. Free or natural convection occurs when the fluid motion is due to a warmer (or cooler) fluid adjacent to a surface. This flow occurs because of the density differences resulting from the temperature variations that result from the heat exchange.

•

Radiation heat transfer is associated with energy that is exchanged by electromagnetic waves that lie in the thermal band of the electromagnetic spectrum. Electromagnetic wave/lengths vary from cosmic waves that have extremely small wave/lengths (known as the low end of the electromagnetic spectrum) to radio waves at the high end. The thermal band is generally considered to encompass the range between 0.38 and 0.76 m (micrometers). In contrast to both conduction and convection, radiation heat transfer requires no medium for its propagation. Indeed, radiant energy exchange between surfaces is at a maximum when the surfaces are separated by a perfect vacuum.

1.5

Engineered Systems and Products

An exhaustive compilation of familiar engineering systems that influence our lives would encompass many pages. The list that follows in Table 1.1 is organized along the lines of general technical sectors, with some subtopics in each sector. These subtopics identify more specific products and processes. Three specific cases will now be considered. 1.5.1 Case I: A Commercial Airliner Figure 1.5 shows a photograph of a commercial airliner that most of us have seen either personally or in the media. Any commercial aircraft is a complicated conglomeration of systems, each of which requires detailed consideration in order to achieve optimal operation. Subsystems of such an aircraft include •

Engine/propulsion

•

Climate control

•

Flight instrumentation

6

Introduction to Thermal and Fluid Engineering TABLE 1.1

Examples of Engineered Systems and/or Products Aerospace • Commercial aircraft • Space systems Environmental control • Greenhouses • Clean rooms • Heat pumps Transportation • Automobiles and trucks • Gas turbines Electronics • Ink jet printers • Chip cooling Public works • Hydroelectricity • Municipal water supplies Manufacturing • Chemical plants • Silicon wafers Power plants • IC engines • Steam and gas turbines • Fuel cells • Nuclear fission •

Flight controls

•

Fuel storage and delivery

•

Engine monitoring and control

•

Hydraulics

FIGURE 1.5 A commercial jet airliner.

The Thermal/Fluid Sciences: Introductory Concepts

7

C1 to C4 Gases Fractionating Column

20°C

C5 to C9 Naphtha

Liquified Petroleum Gas

Chemicals

70°C Fractions Decreasing in Density and Boiling Point

CS to C10 Petrol (gasoline) 120°C

C10 to C16 Kerosine (paraffin oil) 170°C

Petrol for Vehicles

Jet Fuel, Paraffin for Lighting and Heating

C14 to C20 Diesel Oils 270°C

Diesel Fuels

Crude Oil C20 to C50 Lubricating Oil

C20 to C70 Fuel Oil Fractions Increasing in Density and Boiling Point

600°C

>C70 Residue

Lubricating Oils, Waxes, Polishes Fuels for Ships, Factories, and Central Heating Bitumen for Roads and Roofing

FIGURE 1.6 A distillation tower in a petroleum refinery.

•

Communications

•

Navigation systems

The design, manufacture and operation of such a complex system involves a huge array of engineering talents and efforts. Thermal/fluid sciences play roles in every item listed. 1.5.2 Case II: A Chemical Plant Distillation Column The manufacture of chemical products is a major endeavor of contemporary society. Industries that rely on the production and/or conversion of chemicals include pharmaceuticals, agriculture, pulp and paper, plastics, petroleum, and many others. In addition to the manufacture of products from feedstocks, engineers in the chemical industry must meet stringent environmental standards regarding effluents in solid, liquid, and gaseous forms. A component of a petroleum refinery is shown in Figure 1.6. This component is a distillation tower that converts an inlet stream of crude oil to a variety of hydrocarbon products

8

Introduction to Thermal and Fluid Engineering

Resting state.

Rapid heating of 100°C/μsec produces vapor explosion in ink.

Drop formation and bubble expansion.

Bubble collapse begins refill.

Capillary action draws ink into firing chamber.

Meniscus overshoot.

FIGURE 1.7 Drop formation in a thermal ink jet printer. (Adapted Courtesy of the Hewlett-Packard Co., by permission.)

from methane to paraffins. In a petroleum plant, the primary products are fuels for internal combustion engines. The design and operation of such a column involves many components which, in turn, rely on several of the principles within the thermal/fluid sciences. Subsets of the distillation column include •

Extensive piping systems including pumps, valves, and expansion joints

•

Heat exchangers

•

Extensive instrumentation to monitor system temperatures, pressure, and flow rates

•

Controls to adjust operating conditions

1.5.3 Case III: The Thermal Ink Jet Printer Printers, as auxiliaries to personal computers, have become commonplace. In Figure 1.7, we show a representative device that employs a thermal ink jet to deliver ink to a page. The objective of a printer is simple; it is to provide a hard copy image that is clear and informative. The medium employed to impart an image to the paper is ink that may be black or in various colors. The jet of ink is driven by a pressure pulse which, in the thermal process, is provided by the well-controlled and extremely fast formation and collapse of a vapor bubble. The controlled formation and collapse produces a series of ink droplets which, in an aggregate, produce written text, tables, illustrations, and even pictures. The device that accomplishes all of this at an affordable price is a marvelous achievement. The thermal/fluid sciences are vital building blocks to the design, fabrication, and successful operation of these ubiquitous devices. The subsystems of an ink jet printer include: •

Storage and maintenance of an ink at the proper operating conditions

• •

Paper feed operation and control Thermal driver operation

•

Ink delivery and purging systems

•

Fuel storage and delivery

•

Cooling for electronic controls

The Thermal/Fluid Sciences: Introductory Concepts

1.6

9

Historical Development

1.6.1 Thermodynamics Thermodynamics can be traced back almost 2500 years to about 400 BC when Democritus wrote that all matter consists of tiny material bits called atoms. However, developments in thermodynamics were sparse until the beginning of the eighteenth century, although Leonardo da Vinci (1452–1519) stated (about 1500) that air contained two gases; Galileo (1564–1642) approached the concept of temperature (1638) through the use of his thermoscope; Grand Duke Ferdinand of Tuscany invented (1640) the sealed-stem alcohol thermometer; and G. W. Leibnitz (1646–1716) anticipated the first law of thermodynamics by demonstrating (1695) that the sum of kinetic and potential energies remains constant in an isolated mechanical system. The caloric theory held that heat was a subtle elastic fluid possessing mass whose particles repelled each other. It was attracted by ordinary matter in varying degrees and it was indestructible and uncreatable. Moreover, it was either sensible or latent and, in its latent form, it could combine with solids to form liquids and combine with liquids to form gases. The caloric theory was apparently introduced by Joseph Black in 1770, who provided ideas on specific heats and latent heats of phase change and, later, by Black’s student, William Cleghorn, who fortified the caloric theory with several basic postulates. The indestructibility of caloric caused some difficulty. For example, in 1772, John Locke noted that axles were heated by friction at the contact point to the wheels and even Black acknowledged that friction could occur. The caloric theory was further challenged by Count Rumford (Benjamin Thompson, 1753–1814) in his 1798 experiments pertaining to the boring of cannon and Sir Humphrey Davy (1778–1829) who, in 1799, made observations of frictional heating, which indicated that, if indeed there was a caloric fluid, it surely could be created. In 1824, Carnot (1796–1832) formulated the second law of thermodynamics, which appeared to strengthen the caloric theory. His reasoning was that a certain amount of caloric passed through a heat engine in the same manner as water flowed through a turbine and, in doing so, degraded the potential for doing further work. However, it was the quantitative experimental work of Joule (1818–1899), between 1843 and 1849, that led to the law of conservation of energy (or, the first law of thermodynamics). But, in spite of Joule’s work, the caloric theory lasted for a few more years while the argument raged over which of the two laws, Joule’s or Carnot’s, was correct. Indeed, Joule’s theory contradicted Carnot’s, which had been formulated on the basis of caloric theory. The argument persisted until it was carried forward by Helmholtz (1821–1894) in 1847 and by Lord Kelvin (1824–1907) in 1848 and was finally resolved by Clausius (1822–1888) in 1850. Clausius showed that Joule’s work did not require that heat and work be mutually interchangeable under all circumstances and that, because the first and second laws are independent of each other, Carnot’s second law was correct, despite his use of the erroneous caloric theory. The subsequent organizational work of Clausius, Planck (1858–1947), and Poincare (1854– 1912) and extensions of the subject by the American, Gibbs (1839–1903), essentially completed the structure of classical thermodynamics by the beginning of the twentieth century. 1.6.2 Fluid Mechanics Interest in fluid behavior began in the ancient Greek and Roman civilizations. Archimedes (287–212 BC) was the first to express the principles of buoyancy and flotation and Frontinus

10

Introduction to Thermal and Fluid Engineering

(40–103 AD) was apparently the first to provide details on the elaborate water supply systems that were built by the Romans. While there was considerable interest in the design of water conduits, the interest was haphazard until Leonardo da Vinci derived the equation for conservation of mass in onedimensional flow. Coupled with Newton’s (1642–1727) three laws and his law of viscosity pertaining to linear fluids, several of the great mathematicians such as Bernoulli1 (1700–1782), Euler (1707–1783), d’Alembert (1717–1783), Lagrange (1736–1813), and Laplace (1749–1827) provided solutions to the flow problems that arose from the assumption of frictionless flow. Because most flows are particularly vulnerable to the effects of viscosity, engineers rejected the frictionless flow assumption and developed the science of hydraulics, which relied substantially on experimental evidence. The link between theoretical hydrodynamics and experimental hydraulics was established at the end of the nineteenth century. William Froude (1810–1879) and his son Robert (1846–1924) developed the conversion of wave resistance between a model and its prototype. Sir Osborne Reynolds (1842–1912) published the results of his classic pipe experiment (1883), which demonstrated the importance of the dimensionless parameter now known as the Reynolds number. Lord Rayleigh (John W. Strutt, 1842–1919) investigated wave motion, jet instability, and laminar flow analogies. It was the German engineer, Ludwig Prandtl (1875–1953), who introduced the concept of a fluid boundary layer (1904). This concept provided a foundation for the unification of the experimental and theoretical aspects of fluid mechanics. Prandtl proposed that for flow adjacent to a solid boundary, a thin boundary layer, in which frictional effects are very important, is developed. However, outside of the boundary layer, the fluid behaves very much like a frictionless fluid and subsequent investigations by von Karman (1881–1963) and Taylor (1886–1975) provide the basis for the present state of the art in fluid mechanics at the end of the twentieth century. 1.6.3 Heat Transfer Historically, conduction, convection, and radiation, which are generally considered as the three modes of heat transfer, all begin in different places and at different times. Fourier (1768–1830) began working on the problem of heat flow by conduction while he was in Egypt during Napoleon’s invasion and, in 1805, with the help of Biot (1774–1862), he was ready to publish the results of his work. The final publication of his “The Analytical Theory of Heat” was some 20 years later in 1822, 1824, and 1826. These included another of Fourier’s major contributions, the representation of a periodic function in terms of a trigonometric series. In the meantime, Navier (1785–1836) and Poisson (1781–1840) completed the formulation of the viscous field equations that eventually were employed to predict heat convection. Following Lord Kelvin (1824–1907), Stokes (1819–1903) developed a more independent formulation of these equations and these became known as “the Navier-Stokes equations.” It was Maxwell (1831–1879) who set the foundation for the kinetic theory of gases thereby eliminating any further consideration of the caloric theory. Then, Lord Rayleigh (1842–1919) did a considerable amount of work in fluid mechanics that would soon serve all areas of convection heat transfer. While it is observed that convection heat transfer had been driven by developments in fluid mechanics, heat transfer in general and convection heat transfer in particular became a separate discipline as fundamental “real-world” problems began to evolve. 1 These

mathematicians are listed chronologically according to date of birth.

The Thermal/Fluid Sciences: Introductory Concepts

11

In 1904, Prandtl presented his paper proposing the boundary layer and this idea was carried forward by his students, Blasius (1883–1970) and von Karman (1881–1963). Nusselt (1882–1959) was another contributor to the subject of analytical convective heat transfer. His paper in 1909 employed dimensional analysis to prove concepts in natural convection before the basic theory of dimensional analysis by Buckingham (1867–1940) appeared in 1914. In 1884, Boltzmann (1844–1906), who had been working in the area of the kinetic theory of gases to provide a greater capability for the prediction of transport phenomena, turned his attention to thermal radiation. The relationship between emittance and absorptance had been established by Kirchhoff (1824–1887) and, in 1879, Stefan (1835–1893) showed, experimentally, that the heat radiated from a thermally active hot blackbody, should be a function of the fourth power of the absolute temperature. Boltzmann used a well-conceived heat engine argument to show that Stefan’s contention was true and the “fourth power law” is now known as the Stefan-Boltzmann law. It was Wien (1864–1928), a German physicist, who, in 1896, provided his distribution law. This was confirmed by Lord Rayleigh who, in 1900, used classical statistical mechanics to obtain the wavelength distribution of thermal radiation. Finally, Planck (1858–1947), whose explanation of thermal radiation followed in 1901, showed that radiation could only occur in discrete energy levels.

1.7

The Thermal/Fluid Sciences and the Environment

It has been noted that the thermal/fluid sciences form the underpinning for almost all energy transfer and conversion processes. Indeed, they are responsible for many aspects of the quality of life that we all enjoy. However, their utilization is accompanied by an impact on the environment that has become more and more difficult to ignore. The effect of the thermal/fluid sciences on the environment is noticeable in three significant areas. These are global warming due to thermal radiation, the formation of smog due to lift and drag of solid particles, and acid rain, which involves the combustion process. These effects are now briefly considered. 1.7.1 The Ozone Layer The general stratum of the earth’s upper atmosphere in which there is an appreciable ozone, O3 concentration is called the ozone layer or ozonosphere. In this layer (Figure 1.8), the ozone plays an important part in the radiation balance of the atmosphere. The

Absorbs Ultraviolet

Solar Radiation

Ozonosphere

Surface of the Earth FIGURE 1.8 The ozonosphere.

Maximum Ozone Concentration at 20 to 25 km

12

Introduction to Thermal and Fluid Engineering Water Level

Water Flow

Water Flow Water Level

FIGURE 1.9 Artist’s conception of a hydroelectric trubine employed in a hydroelectric power plant.

layer lies roughly between 10 and 50 km above the surface of the earth with maximum ozone concentration at about 20 to 25 km. The ozone layer transmits most of the oncoming solar radiation incident upon it but, thankfully, provides a barrier to harmful ultraviolet radiation. The temperature at the surface of the earth increases during daylight hours and the earth cools down at night by radiating part of the energy to deep space via infrared radiation. Water vapor, carbon dioxide and traces of methane, and the oxides of nitrogen inhibit the nocturnal radiation. These are called greenhouse gases, which lead to a phenomenon called the greenhouse effect. The greenhouse effect keeps the earth warm and makes life, as we know it, possible. However, excessive amounts of these greenhouse gases causes some of the heat reradiated by the earth to be trapped by these gases. This causes the average temperature of the earth to increase and the climate at some localities to undergo an undesirable change. The phenomenon is usually referred to as global warming and can cause severe changes in weather patterns and the melting of ice at the poles. The seriousness of global warning has resulted the United Nations forming a committee to deal with climate changes due to global warming and world summits were held in Rio de Janiero (1992) and Kyoto (1997) to set goals for the reduction of CO2 and other greenhouse gases. These goals can be assisted by resorting to the use of hydroelectric (Figure 1.9), wind (Figure 1.10), and geothermal energy sources. 1.7.2 Smog Smog is a brown and/or yellow haze that exists as a stagnant air mass and surrounds congested and populated areas on warm or hot summer days. It is composed of carbon monoxide, volatile organic compounds such as butane and other hydrocarbons, dust and soot in particulate form, and ground level ozone. Carbon monoxide is an odorless and colorless poisonous gas that is emitted by motor vehicles. It can combine with red blood cells that actually carry oxygen, decreasing the supply of oxygen to the brain and other organs, thereby slowing body reactions and reflexes. At high concentrations, an excess of carbon monoxide in the environment can be fatal. On the other hand, ground level ozone, which should not be confused with the ozone layer that exists in the atmosphere, is a pollutant that has significant adverse health effects. It can cause smartness and irritation to the eyes, shortness of breath, general fatigue, headaches,

The Thermal/Fluid Sciences: Introductory Concepts

13

Wind

Swivel Wind

FIGURE 1.10 Artist’s conception of the use of wind power.

and nausea. It damages the air sacs in the lungs causing temporary problems such as asthma. It is formed when hydrocarbons and the oxides of nitrogen react in the presence of sunlight. Although ground level smog and ozone form in congested urban areas with heavy traffic (the Los Angeles freeways, for example) prevailing winds can transport them to other locations. This makes the production of smog a global rather than a local problem. 1.7.3 Acid Rain Fossil fuels such as coal, gasoline, and diesel oil contain an amount of sulfur that varies from just a trace to an appreciable amount. During the combustion process in motor vehicles and power plants where nitrogen is a major ingredient of the air used, oxides of sulfur, and nitrogen are produced (Figure 1.11). In the presence of water vapor, which is also contained in the products of combustion, the oxides of sulfur and nitrogen form droplets of sulfuric and nitric acid. These droplets are washed from the atmosphere and form what is known as acid or acid rain. The soil is capable of neutralizing certain amounts of this acid rain but the amounts produced by power plants using high-sulfur-laden fuels may exceed this capability. The result may be the deterioration and eventual destruction of forests because the leaves and root systems of the trees and shrubs absorb these acids. Indeed, lakes and rivers may become too acidic for fish to spawn and exist.

S + O2 SO2 + H2O 2H2SO3 + O2 N + O2 NO2 + H2O 2H2NO3 + O2

FIGURE 1.11 The combustion of sulfur and nitrogen in air.

SO2 H2SO3

(Sulfurous acid)

2H2SO4 (Sulfuric acid) NO2 H2NO3 (Nitrous acid) 2H2NO4 (Nitric acid)

14

Introduction to Thermal and Fluid Engineering

The problem of acid rain has been recognized and legislation that prohibits the use of high-sulfur-laden coal now exists.

1.8

Summary

In this chapter, we have examined the component disciplines that comprise the thermal/fluid sciences. These components are thermodynamics, fluid mechanics, and heat transfer. The fundamental laws, which are the building blocks of these disciplines, have been introduced and discussed in general terms. Important new properties have also been introduced. In this chapter, we have also included examples of engineered systems whose development and operation involve the thermal/fluid sciences to a significant degree. The chapters to follow will address thermodynamics, fluid mechanics, and heat transfer in considerable detail. Numerous illustrative examples will be employed to illustrate techniques for applying the basic laws.

2 Thermodynamics: Preliminary Concepts and Definitions

Chapter Objectives •

To briefly introduce the subject of thermodynamics.

To provide precise definitions of some of the working terms used in a study of thermodynamics. • To consider the dimensions and units that pertain to thermodynamics. •

•

To examine density and its related properties.

•

To define pressure and consider how it is measured.

•

To define temperature and to present the zeroth law of thermodynamics.

•

To outline a problem-solving methodology.

2.1

The Study of Thermodynamics

When most people think about thermodynamics, they think about the transfer of energy and the utilization of such energy transfer for the useful production of work. This often leads many engineering students in fields such as computer science and electrical or civil engineering to wonder why this particular subject is relevant to them. In reality, thermodynamics deals with much more than the study of heat or energy transfer and the development of work. Indeed, it deals with virtually all aspects of our lives, from the combustion processes that run our automobiles and produce our electric power in power plants to the refrigeration cycles that cool our beer, from the cryogenic pumping of liquids and gases in space to the distillation processes used to produce the gasoline that runs our automobiles. Thermodynamics is important to electrical engineers so that they can better understand that the limiting factor in the microminiaturization of electronic components is the rejection of heat. It is important to civil engineers because a knowledge of thermal expansion and thermal stresses is requisite to the design of structures and to the computer scientists who need to thoroughly understand the systems that they are trying to model and develop. Many engineering students have heard the rumors about thermodynamics that say it is a difficult subject, it is a great leveler in that taking a course in it will destroy one’s grade point average and that it is something to be avoided until it becomes a matter of taking it or not graduating. Be assured that the authors have been there. All of this is, of course, not 15

16

Introduction to Thermal and Fluid Engineering

true and if one keeps an open mind, a study of thermal and fluid sciences in general and thermodynamics in particular can be quite a rewarding experience. Common terms like work, reversibility, and system all have very well-defined meanings in thermodynamics and we will study them. Moreover, we will also study a whole host of terms that are completely unfamiliar to the student who is not pursuing a curriculum in mechanical or chemical engineering. Fortunately, however, everyone knows a little something about thermodynamics, which has been learned through everyday experience, that a ball rolls downhill, that heat moves from a region at high temperature to one at low temperature, and that wind-up toys will just not run on their own. Thermodynamics is with us in our everyday lives and, through its study, we can better relate to the world around us. The first law of thermodynamics can help us understand the need for conservation of our natural resources and the second law of thermodynamics can give us an understanding of why it is so difficult to separate gases after they mix and why your sock drawer is always such a mess. And, to be sure, the property of entropy can help us better understand the vast extent and operation of systems and devices in our universe. The harnessing of available energy and the transformation of this energy to a usable form has long been a goal of society. Indeed, energy has driven society, in a little more than a century, from muscular effort and the horse and buggy to the modern day automobile, supersonic transport, modern tools and appliances, heating, air conditioning, and the exploration of space. Thermodynamics provides the basic principles of energy transfer. However, because the development of particular sources of energy is becoming increasingly expensive, thermodynamics is also concerned with the efficiency of energy utilization. Moreover, thermodynamics is concerned with the environmental impact of various energy conversion alternatives. These are the reasons why we study thermodynamics. Our first step will be the definition of some of the working terms.

2.2

Some Definitions

2.2.1 Systems A system is any portion of the universe that is chosen for thermodynamic analysis. It can be real, imaginary, fixed, or movable and it is separated from its surroundings or environment by its boundaries (Figure 2.1). If neither mass nor energy crosses its boundaries, it is said Surroundings

System

System Boundary FIGURE 2.1 A system is separated from its surroundings by its boundaries.

Thermodynamics: Preliminary Concepts and Definitions

No Mass In

×

Closed System

Energy In

17

Energy Out ×

No Mass Out

FIGURE 2.2 A closed system or control mass. Energy may flow in or out but mass may not.

to be an isolated system. In a closed system or control mass, only energy (and no mass) may be transferred across the system boundaries (Figure 2.2). If both mass and energy are transmitted across the system boundaries, the system is said to be an open system or control volume (Figure 2.3). If the properties of the system (to be discussed shortly) are everywhere the same, the system is considered uniform and if the properties do not change with time, the system is said to be at steady state. 2.2.2 Fluids A fluid is a substance that deforms continuously under the action or the influence of a shear force. This characteristic allows it to assume the shape of its container. Both liquids and gases are fluids. 2.2.3 Substances A substance is a tangible material that occurs in macroscopic amounts and a pure substance is one that is uniform and homogeneous in chemical structure. A pure substance may exist in a single phase (solid, liquid, or vapor), as a two-phase (solid-liquid, liquid-vapor, or solid-vapor) mixture, or as a three-phase (solid-liquid-vapor) mixture. In a two-phase or three-phase mixture, the phases are separated by phase boundaries (Figure 2.4). We may note that the vapor phase is frequently referred to as the gaseous phase, which brings us to the difference between a gas and a vapor. We find that the terms gas and vapor are often used interchangeably. However, they may be distinguished from one another by recognizing that a substance which, at standard conditions of temperature and pressure, normally exists in the gaseous phase (oxygen and nitrogen, for example) is normally called a gas. On the other hand, if a substance at standard conditions of temperature and pressure normally exists as a liquid (water, for example), it can produce a vapor.

Mass In Energy In

Open System

Energy Out Mass Out

FIGURE 2.3 An open system or control volume. Both energy and mass may flow in or out and the mass flows may contain energy.

18

Introduction to Thermal and Fluid Engineering

Phase Boundary

Steam

Liquid Water FIGURE 2.4 A two-phase system containing steam or vapor and liquid water. The two phases are separated by the phase boundary.

2.2.4 State and Properties The state of a system is the condition in which the system exists. A thermodynamic property is any observable characteristic that describes the state of the system. Properties that depend on the mass of the system and hence the extent of the system, such as volume (designated by V) and energy (designated by E), are called extensive properties. Properties that do not depend on the extent of the mass in the system such as pressure (designated by P) and temperature (designated by T) are called intensive properties. We note that both extensive and intensive properties are designated by capital letters. Extensive properties, when given on a per unit mass basis are called specific properties and are designated by a lower case letter (Figure 2.5). Examples of specific extensive properties are specific volume, v = V/m, and specific energy, e = E/m. 2.2.5 Processes A simple substance is a substance whose state is determined by two independent intensive properties that are independently variable (Figure 2.6). We observe that the density,  = m/V is not an independent intensive property because it is the reciprocal of the specific volume, v. We also see that, because the state of a system is determined by its properties, whenever any property changes, its state changes, and the system is said to have undergone a process. A sequence of processes that returns a system to its initial state is called cycle (Figure 2.7) and we note that the change in the values of a property depends only on the end states of the process and is independent of the process itself. Therefore, the properties of a system that undergoes a cycle are the same at the initial and final states.

Boundary between Subsystems

V1 E1 T1 P1 V2 E2 T2 P2

Subsystem-1

Subsystem-2

FIGURE 2.5 Two subsystems separated by a boundary. The subsystems contain substances having the extensive properties of volume, V, and energy, E, and the intensive properties of temperature, T, and pressure, P.

Thermodynamics: Preliminary Concepts and Definitions

19

P

Pressure

P2V2

P1V1 V

Volume

FIGURE 2.6 A P-V diagram of a process in which the state of the system changes because the system properties change.

2.2.6 Thermodynamic Equilibrium 2.2.6.1 Requirements for Thermodynamic Equilibrium A system is said to be in thermodynamic equilibrium when it cannot change its state without some outside influence, that is, when there is no tendency of the system to change spontaneously. In order to achieve thermodynamic equilibrium, mechanical, thermal, chemical, and phase equilibrium must all exist individually and simultaneously: •

Mechanical equilibrium requires a condition of balance between all opposing forces, moments, and other influences.

•

Thermal equilibrium requires all parts of the system to be at the same temperature (Figure 2.8).

•

Chemical equilibrium requires that the system show no tendency to undergo further chemical reaction.

•

Phase equilibrium requires that the system be either in a single phase or in a number of phases that show no tendency to change their condition when the system is isolated from its surroundings.

A system that is in equilibrium (Figure 2.8) is said to be in an equilibrium state. P

Pressure

P2V2

P1V1 Volume

V

FIGURE 2.7 A P-V diagram of a cycle indicating the properties of the system at both the initial and the final states. Both the initial and final states are indicated by P1 and V1 . Here, P2 and V2 are at some intermediate state.

20

Introduction to Thermal and Fluid Engineering Thermal Barrier

40°C

60°C

80°C

60°C

(a)

(b)

Thermal Contact

FIGURE 2.8 Two systems (a) not in thermal equilibrium and (b) in thermal equilibrium.

2.2.6.2 Quasi-Equilibrium or Quasi-Static Processes Regardless of the process, a system can proceed from one equilibrium state to another only if the system is perturbed, that is, only if some external force or other influence acts upon the system. At the end of the process, another, but different, equilibrium state (which also isolates the system from its surroundings) is obtained. If the process occurs at a finite rate, deviations from equilibrium conditions may become significant. However, the process may be thought of as a series of infinitesimal processes, each of which achieves an equilbrium state at its conclusion. Under these circumstances, we see that the entire process is a sum of a series of quasi-equilibrium or quasi-static processes (Figure 2.9) in which the deviation from thermodynamic equilibrium is not only negligible but infinitesimal. 2.2.7 Statistical and Classical Thermodynamics The microscopic approach to the study of thermodynamics involves individual molecules and their internal structure which may include individual atoms, electrons, and nuclei. Because the number of these particles is huge, statistical techniques using the laws of probability can be employed to determine the most probable distribution of particles between momentum and energy states. When these statistical techniques are employed, the microscopic viewpoint is then described by statistical thermodynamics or statistical dynamics. The macroscopic approach to the study of thermodynamics, which we use in this book, involves quantities of matter that are finite in size. The detailed molecular or atomic structure of matter is ignored and the macroscopic approach is adopted in the study of classical thermodynamics.

W1 W2 W3 W4 W5 W6

Compressed Spring

FIGURE 2.9 A platform with several weights held in place by a compressible spring. As each weight is removed, the platform rises in a series of quasi-equilibrium or quasi-static processes.

Thermodynamics: Preliminary Concepts and Definitions

21

TABLE 2.1

Primary Dimensions and Units for the SI System

2.3

Dimension or Quantity

SI Unit and Symbol

Mass Length Time Temperature Amount of substance Luminous intensity Electric current

kilogram (kg) meter (m) second (s) Kelvin (K) mole (mol) candela (Cd) ampere (A)

Dimensions and Units

2.3.1 Introduction Primary dimensions refer to a rather small group of physical quantities such as mass, time, length, and temperature that may be employed to describe the nature of a physical system. Secondary dimensions such as pressure, velocity, and area are measured in terms of the primary dimensions. A system of units is used to provide arbitrary magnitudes and give names that are assigned to the dimensions. The units are adopted as standards of measurement and are referred to as base units. 2.3.2 The SI (Syst`eme Internationale D’unit´es) System The SI system of units is an extension of the metric system and has been adopted in many countries as the only legally acceptable system. SI units are divided into three classes that are the base units, the derived units, and some special units. The primary dimensions used in a study of the thermal and fluid sciences along with their base units and symbols embrace the first five entries in Table 2.1. There are “standards” for all of the units shown in Table 2.1. For example, the standard for the second is the duration of 9,192,631,770 periods corresponding to the transition states between two levels of the ground state of the cesium-133 atom. Examples of derived units expressed in terms of the base units are given in Table 2.2 and examples of derived units with special names are given in Table 2.3. TABLE 2.2

Examples of SI Derived Units Expressed in Terms of the Base Units Dimesion or Quantity Area Volume Velocity Acceleration Density Specific volume Current density

SI Unit Name and Symbol square meter (m2 ) cubic meter (m3 ) meter per second (m/s) meter per second squared (m/s2 ) kilogram per cubic meter (kg/m3 ) cubic meter per kilogram (m3 /kg) ampere per square meter (A/m2 )

22

Introduction to Thermal and Fluid Engineering TABLE 2.3

Examples of SI Derived Units with Special Names Name and Symbol

Dimension or Quantity Force Pressure Energy, heat, work Work Electrical charge Electrical potential Electrical resistance Magnetic flux Electrical inductance

Expression in Terms of Other Units

newton, N pascal, Pa joule, J watt, W coulomb, C volt, V ohm,  weber, Wb henry, L

m-kg/s2 N/m2 m2 -kg/s2 J/s A-s W/A V/A V/s Wb/A

Expression in Terms of Base Units kg/m-s2 m2 -kg/s3 A-s m2 -kg/s3 -A m2 -kg/s3 -A2 m2 -kg/s2 -A m2 -kg/s2 -A2

The mole is defined as the molecular weight of a substance expressed in the appropriate mass unit. For example, a gram-mol (gmol) of hydrogen contains 2.016 grams (g) and a kilogram-mol (kgmol) of hydrogen contains 2.016 kilograms (kg). The number of moles of a substance, n, is related to its mass, m and the molecular weight, M, by the simple expression n=

m M

(2.1)

In the SI system, force is a secondary dimension because the unit of force is a derived unit. Table 2.3 shows that the unit of force is the newton, N, which can be obtained from Newton’s second law, F ∝ ma as 1 N = (1 kg)(1 m/s2 ) = 1 kg-m/s2 and we see that the constant of proportionality is unity or 1

N-s2 =1 kg-m

(2.2)

This allows Newton’s law to be written as F = ma . Because pressure is force per unit area, P = F /A, the unit of pressure, the pascal, can be expressed in terms of 1 Pa = 1 N/m2 = (1 kg-m/s2 )(1 /m2 ) = 1 kg/m-s2

(2.3)

Classically, a work interaction or work, W, is defined as the dot or scalar product of a force vector, F, and a displacement vector, dx, as indicated in Figure 2.10. W = F · dx = Fdx cos  where  is the angle between the line of action of the force, F, and the displacement direction, x. Here W represents an inexact differential which, along with the direction of the work interaction, is discussed in detail in Chapter 3. In this book, a work interaction or work will be represented by W = F x and we recognize that the unit of work is the joule (J) defined as 1 J = 1 N-m

(2.4)

Thermodynamics: Preliminary Concepts and Definitions

23

F θ

W = mg

dx x FIGURE 2.10 A force applied at an angle  with the horizontal. The weight moves in the horizontal direction indicated by the length coordinate, x, and the differential amount of work done is given by the dot or scalar product of the vector force, F. and the displacement, x, or W = F · dx.

and because power, P, is the rate of doing work, the unit of power is the watt (W) P=

W ˙ = 1 J/s = 1 N-m/s =W dt

(2.5)

Finally, we note that the weight of a body is equal to the force of gravity on the body. Hence, weight always refers to a force and, in the SI system, this force is always in newtons. The mass of the body is related to the weight via W = mg

(2.6)

where g is the local gravitational acceleration, which has a mean sea level value of g = 9.807 m/s2 and is a function of location. This shows that the weight of a body may vary while the mass of the body is always the same (Figure 2.11). 2.3.3 The English Engineering System The English Engineering System of units is often used in the United States. This system adds the force in lbf as a primary dimension as indicated in Table 2.4. Here we find the

F=1N

m = 1 kg

a = 1 m/s2

m = 1 kg g = 9.807 m/s2 W = 9.807 N FIGURE 2.11 The weight of a body may vary but the mass of the body is invariant. Here, in the SI system of units, gc = 9.807 m/s2 .

24

Introduction to Thermal and Fluid Engineering TABLE 2.4

Primary Dimensions and Units for the English Engineering System Dimension or Quantity

Unit and Symbol

Mass Length Time Temperature Amount of substance Luminous intensity Electric current Force

pound-mass (lbm ) foot (ft) second (s) degree Rankine (◦ R) mole (mol) candle ampere (A) pound-force (lbf )

pound used as both the unit of mass (lbm ) and the unit of force (lbf ). This leads to more than just a little confusion when this system is employed for two reasons: •

The use of the symbol, lb, is not permitted and use must always be made of the subscript to designate lbm or lbf .

•

Because there are six primary dimensions, Newton’s law F ∝ ma must be written with a proportionality constant (Figure 2.12) F =

ma gc

(2.7)

where gc is a proportionality constant. The force required to accelerate a mass of 1 lbm at a rate of 32.174 ft/s2 is designated as 1 lbf . Thus, 1 lbf =

(1 lbm )(32.174 ft/s2 ) gc

or gc = 32.174

lbm -ft lbf -s2

(2.8)

Thus, Newton’s law must be written as F =

F = 1 lbf

ma 32.174

m = 32.174 lbm

(2.9)

a = 1 ft/s2

m = 1 lbm g = 32.174 ft/s2 W = 1 lbf FIGURE 2.12 Weight and mass in the English system of units require the use of a proportionality constant, gc = 32.174 m/s2 for their conversion.

Thermodynamics: Preliminary Concepts and Definitions

W = 200 N

25

g = 9.45 ft/s2

(a)

W=?

g = 2.25 ft/s2

(b) FIGURE 2.13 Data for Example 2.1.

It is important to remember that the SI system of units requires this conversion factor between force and mass, gc but, because its value is unity, it is often omitted. In this book, we will employ the SI system exclusively.

Example 2.1 A metal ingot weighs (Figure 2.13) 200 N at a location where the acceleration of gravity is 9.45 m/s2 . Determine its mass and weight when the acceleration of gravity is 2.25 m/s2 .

Solution Assumptions and Specifications 1. The metal ingot is the system. 2. Two values of the acceleration of gravity are specified. Here, the mass can be computed using Equation 2.6 m=

W 200 N = = 21.16 N-s2 /m g 9.45 m/s2

and because 1 kg = 1 N-s2 /m, the mass is m = 21.16 kg ⇐ The mass of the ingot is invariant but its weight changes as the acceleration of gravity changes. Hence, when g = 2.25 m/s2 W = mg = (21.16 kg)(2.25 m/s2 ) = 47.61 kg-m/s2 = 47.61 N ⇐ 2.3.3.1 The British Gravitational System In Section 2.3.3, we observed that the lbm is the preferred unit of mass in the English Engineering System. Another unit of mass is the slug, which is defined as the quantity of mass that could be accelerated at 1 m/s2 when it is acted upon by a force of 1 lbf . With the slug designated as the preferred mass unit, we have the British Gravitational System with all of the entries in Table 2.4 remaining the same except for the unit of mass that is changed from the lbf to the slug. In this case, Newton’s second law reads 1 lbf = (1 slug)(1 m/s2 ) = 1 slug-ft/s2 which shows that the relationship between the slug and the pound mass is 1 slug = 32.174 lbm

26

Introduction to Thermal and Fluid Engineering TABLE 2.5

Standard SI Unit Prefixes Factor 10−12 10−9 10−6 10−3 10−2

Prefix

Symbol

pico nano micro milli centi

p n  m c

Factor

Prefix

Symbol

102

hecto kilo mega giga tera

h k M G T

103 106 109 1012

2.3.4 SI Unit Prefixes Because it is often necessary to work with very large or very small values of the quantities in the SI system, standard prefixes have been provided to simplify the usage. These are listed in Table 2.5. With these prefixes in hand, 1,000,000 watts become 1 MW, 1000 grams become 1 kg and 0.001 m becomes 1 mm.

2.4

Density and Related Properties

The density, , of a substance is defined as its mass per unit volume =

m V

(2.10)

and the reciprocal of the density is the specific volume, v v=

1 V =  m

(2.11)

The specific gravity of a substance is the ratio of its density to the density of water at some reference temperature  SG = (2.12) H2 O and we note that, because specific gravity is a ratio, it has no units or dimensions. The density of water at 25◦ C can be taken as H2 O = 997.11 kg/m3 and, because the density of mercury is 13,560 kg/m3 , its specific gravity can be taken as 13.6. The specific weight, , of a substance is its weight per unit volume =

W mg = = g V V

(2.13)

It is sometimes necessary to express the specific volume and other specific quantities on a per mole basis. Thus, for the specific volume, v¯ = Mv

(2.14)

Thermodynamics: Preliminary Concepts and Definitions

111.7 kg of Red Gage Oil

27

V = 0.08 m3

FIGURE 2.14 A quantity of gage oil that occupies a certain volume.

where the molal specific volume in m3 /kgmol is the quantity carrying the overbar and where M is the molecular weight of the substance in kg/kgmol. Values for the molecular weight for various substances are listed in Table A.1 in Appendix A.

Example 2.2 As shown in Figure 2.14, 111.7 kg of red gage oil occupies a volume of 0.08 m3 . Determine its density, specific volume, specific gravity, and its specific weight at sea level.

Solution Assumptions and Specifications

1. The red gage oil is the system. 2. The mass and volume are specified.

The density is given by Equation 2.10 =

m 111.7 kg = = 1396.3 kg/m3 ⇐ V 0.08 m3

and the specific volume is the reciprocal of the density v=

1 1 = = 7.16 × 10−4 m3 /kg ⇐  1396.3 kg/m3

With the density of water taken as 997.11 kg/m3 , the specific gravity is computed using Equation 2.12 SG =

 H2 O

=

1396.3 kg/m3 997.11 kg/m3

= 1.40 ⇐

Finally, we find that because the specific weight is related to the density by the acceleration of gravity, which at sea level is g = 9.807 kg/m3  = (1396.3 kg/m3 )(9.807 m/s2 ) = 13, 693.5 kg/m2 -s2 Because 1 kg = 1 N-s2 /m, then 1 kg/m2 -s2 = 1 N/m.3 Thus,  = 13, 693.5 N/m3 ⇐

28

2.5

Introduction to Thermal and Fluid Engineering

Pressure

2.5.1 Gage, Absolute, and Vacuum Pressure Pressure is a type of stress that is exerted uniformly in all directions. The pressure, P, is defined as the normal force per unit area exerted on the confining surface of a fluid system P=

d Fn dA

(2.15)

and is created by bombardment of the surface by the fluid molecules. Here, the differential area, d A, is the smallest area for which the effects are the same as for a continuous medium or continuum. The unit of pressure is N/m2 and is called pascal. Because the pascal is a small quantity, kPa = 103 Pa and MPa = 106 Pa are often used in thermodynamics. If we consider zero pressure at a datum level, the actual pressure at a given position in the fluid system can be considered as its absolute pressure. However, most pressure measuring devices indicate a gage pressure that is the difference between the absolute pressure and the atmospheric pressure taken as  1 standard atmosphere =

101, 325 Pa 760 mm of Hg at 20◦ C

We also note that a frequently used unit of pressure is the bar 1 bar = 105 Pa = 100 Pa The gage pressure is the difference between the absolute pressure in the system and atmospheric pressure Pgage = Pabs − Patm

(2.16a)

For pressures below atmospheric, the term vacuum indicates the difference between the atmospheric pressure and the absolute pressure Pvac = Patm − Pabs

(2.16b)

These relationships are summarized in Figure 2.15 where we see that the datum for absolute zero pressure is a perfect vacuum. 2.5.2 Measuring Pressure Pressure may be measured by Bourdon tube gages, pressure transducers, strain gages, piezoelectric sensors, or by manometers (Figure 2.16), which are devices that measure pressure differences in terms of the height or length of a column of liquid such as water, mercury or, what is commonly known as gage oil. A manometer that measures atmospheric pressure is called a barometer (Figure 2.17), although not all barometers are manometers. Manometers are discussed in detail in Chapter 12.

Thermodynamics: Preliminary Concepts and Definitions

29

Gage Pressure

Vacuum Pressure

Absolute Pressure

Absolute Pressure Atmospheric Pressure

FIGURE 2.15 The relationship between absolute, gage, atmospheric, and vacuum pressure.

2.6

Temperature and the Zeroth Law of Thermodynamics

Temperature is that property of a substance or object that determines the direction of heat flow when the object is placed into contact with another object. Unfortunately, we find that the words “property” and “direction of heat flow” are insufficient for the precise evaluation of temperature that is required in our study of thermodynamics and, as a result, it is often advantageous to think of temperature as a measure of molecular activity. Consider body A at temperature TA, which is brought into intimate physical contact with body B at temperature TB . If both bodies are isolated from the surroundings with TA > TB , heat will flow from body A to body B and after a period of time of sufficient duration has elapsed, there will be no further heat flow from body A to body B. At this time, both bodies will have reached a state of thermal equilibrium (Figure 2.18) and both may be presumed to be at the same temperature. This serves to define equality of temperature and leads to the statement of the zeroth law of thermodynamics.

Path

Gas at Pressure, P

L

Manometer Liquid FIGURE 2.16 Manometer.

30

Introduction to Thermal and Fluid Engineering

760 mm

29.92 in

Mercury FIGURE 2.17 A mercury barometer at the standard atmospheric pressure of 29.92 in (760 mm) of mercury.

If body A is in thermal equilibrium with body C and if body B is also in thermal equilibrium with body C, then body A and body B must be in thermal equilibrium with each other. The zeroth law of thermodynamics forms the basis of temperature measurement or thermometry. A thermometer is a device that has at least one property that varies with temperature. The property itself is referred to as a thermometric property and the substance that displays this property is called a thermometric substance. In the common “mercuryin-glass” thermometer (Figure 2.19), the mercury is the thermometric substance and the coefficient of expansion that causes a change in volume of the mercury is the thermometric property. Thermometers are not the only temperature measuring devices. Thermometry, which is the art of temperature measurement also makes use of changes of pressure in a gas confined at constant volume, changes in electrical resistivity (the thermistor), changes in electrical potential (the thermocouple), and optical changes (the radiometer). We note that the third body considered in the zeroth law may be employed as a thermometer that can be calibrated by bringing it into thermal equilibrium with a body at known temperature such as a mass of melting ice. The absolute temperature scale in the SI system is the Kelvin scale and the unit of temperature is kelvin (K without the degree symbol). Its reference state value is the triple point of water, which is the point where a state of phase equilibrium exists among the three

System A At TA

System C At TC

System B At TB

FIGURE 2.18 Three systems in close proximity. If TA = TB = TC , the systems are said to be in thermal equilibrium.

Thermodynamics: Preliminary Concepts and Definitions

31

FIGURE 2.19 Mercury-in-glass thermometer.

phases: ice (solid), water (liquid), and steam (vapor). This point occurs at a pressure of one atmosphere and a temperature of 273.16 K. The freezing point of water, called the ice point, is the state of phase equilibrium between ice and water at one atmosphere. It is set at a point 0.01 K lower at 273.15 K. The Celsius temperature scale (formerly known as the centigrade temperature scale) has 0◦ C (note the use of the degree symbol) as the ice point. Hence, ◦

C = K − 273.15◦

and we will find enough accuracy in the relationship ◦

C = K − 273◦

(2.17)

The boiling point of water, which is the state of equilibrium between steam and liquid water at 1 atmosphere, is set 100 K higher at 373.15 K. The difference between the ice point and the boiling point of water is 100◦ C. The temperature scales used in the English system of units are the Rankine and the Fahrenheit scales. We know that in the Fahrenheit scale, the ice point is taken as 32◦ F and the difference between the ice point and the boiling point is 180◦ F. Following the proportion, 180/100 = 1.80 between the Fahrenheit and the Celsius scales, the Rankine temperature (◦ R) is defined as 1.80 times the temperature in kelvin ◦

R = 1.80 K

(2.18)

which puts the triple point of water at 491.69◦ R and the ice point of water at 459.67◦ F. The four temperature scales are shown in Figure 2.20. From the foregoing, we can list the conversions between the Celsius and Fahrenheit scales as ◦

C=

5 ◦ ( F − 32◦ F) 9

(2.19a)

32

Introduction to Thermal and Fluid Engineering T

K

°C

°R

°F

H2O Boiling Point

373.15

100

671.67

212

H2O Ice Point

273.15

0

491.67

32

Absolute Zero

0.00

–273.15

0

–459.67

FIGURE 2.20 Kelvin, Celsius, Rankine, and Fahrenheit temperature scales and the relation between them.

and ◦

F=

9◦ C + 32◦ C 5

(2.19b)

We can incorporate both of these into the single expression ◦

◦ C F − 32◦ F = 5 9

2.7

(2.19c)

Problem-Solving Methodology

Simplification of a problem with provision of appropriate fundamental principles is called modeling and the result is in the form of a mathematical model or merely, a model. If the model does not require a computer solution, certain steps are needed in order to come to an expeditious solution involving the following steps: •

Determine the purpose of the problem.

•

State what is given with the realization that some of the information may eventually prove to be unnecessary.

•

Ascertain what is to be found.

•

Make and list any assumptions that you may need in order to simplify the problem.

•

If required, draw and label a sketch identifying all known values.

•

Consider what fundamental principles may be needed.

•

Carefully consider the strategy for the solution and check alternative approaches to determine the most expeditious solution procedure.

•

Label each step of the solution.

•

Be sure that each answer is properly identified and has the proper units.

•

Make sure that the solution of a “real-world” problem is a “real-world” solution.

•

Provide any comments deemed to be necessary to support your answer.

Thermodynamics: Preliminary Concepts and Definitions

33

Observe that the foregoing list is presented as a guide and some judgment as to what to include may be required.

2.8

Summary

In this chapter we have presented a discussion of the working terminology that is germane to a study of thermodynamics. This discussion was based on the definitions of systems, fluids, substances, states and properties, and processes. In addition, the various types of equilibrium were listed and quasi-equilibrium and quasi-static processes were considered. The difference between a control mass and a control volume is fundamental and these definitions have been provided in Section 2.2.1. •

A closed system, often referred to as a control mass, is a system in which energy may be transferred across its boundaries but there is no transfer of mass across the boundaries from or to the surroundings.

•

An open system or control volume, is a system in which both energy and mass may cross the system boundaries.

The difference between the control mass and control volume system formulations is in the transfer of mass across the system boundaries. The mass that is transferred may contain energy. A discussion of the SI system of units was provided and it was shown that the primary dimensions having a bearing on thermodynamics were mass, length, time, the amount of substance, and temperature. Force was shown to be a secondary dimension because the unit of force is a derived unit. The English engineering system was then considered and it was shown that, in this system, both mass and force are primary dimensions. The proportionality constant, gc , was then developed and it was shown that in the SI system gc = 1

kg-m N-s2

and in the English engineering system gc = 32.174

lbm -ft lbf -s2

Density, specific volume, specific volume, specific gravity, specific weight and gage, absolute and vacuum pressure were then discussed. After an exposition of the zeroth law of thermodynamics including the definition of temperature scales and the conversion between them, the chapter concluded with some thoughts on the methodology of problem solving.

2.9

Problems

Weight, Mass, and Newton’s Law 2.1: A force of 80 N is applied to a mass of 9 kg. Determine the acceleration of the mass. 2.2: Determine the weight of a 40-kg mass in a location where g = 8.25 m/s.2

34

Introduction to Thermal and Fluid Engineering

2.3: An object weighs 400 N on the surface of the earth where g = 9.81 m/s.2 Determine its weight on a planet where the acceleration of gravity is 1.25 m/s.2 2.4: A system has a mass of 8 kg and is subjected to an external vertical force of 160 N. If the local gravitational acceleration is 9.47 m/s2 and frictional effects may be neglected, determine the acceleration of the mass if the external force is acting (a) upward and (b) downward. 2.5: An inhabitant of another planet weighs 150 N on a spring-type scale in his planet atmosphere where g = 1.85 m/s.2 If he appears at a location on earth where the local gravitational acceleration is 9.68 m/s, 2 determine (a) his mass on his planet, (b) his mass on earth, and (c) his weight on a spring scale. 2.6: Given that the acceleration of gravity decreases at a rate of 3.318 × 10−6 /s2 for every meter of elevation, take the acceleration of gravity at sea level as 9.807 m/s2 and determine the weight of a 100-kg mass in Denver, Colorado where the altitude is 1609.3 m. 2.7: Given that the acceleration of gravity decreases at a rate of 3.318 × 10−6 /s2 for every meter of elevation, take the acceleration of gravity at sea level as 9.807 m/s2 and determine the altitude where the weight of an object is decreased by 5%. 2.8: A 2-kg mass is “weighed” with a beam balance at a location where g is 9.75 m/s2 . Determine (a) its weight and (b) its weight as determined by a spring scale that reads correctly for a standard gravitational acceleration of 9.81 m/s.2 2.9: An astronaut who weighs 1962 N at Cape Canaveral where g = 9.81 m/s2 goes to a planet where g = 5.25 m/s2 . While on the planet, he gathers a bag of rocks weighing 105 N to be brought back as samples. Upon his return to earth, it is found that the rocks and the astronaut together weigh 2011 N. How much weight did the astronaut lose during the mission. 2.10: An object that occupies a volume of 0.625 m3 weighs 3920 N where the acceleration of gravity is 9.80 m/s2 . Determine the mass of the object and its density. Density and Related Properties 2.11: A cylindrical drum that is 60 cm in diameter and 85 cm high is completely filled with a fluid whose density is 1040 kg/m3 . Determine (a) the total volume of the fluid, (b) the total mass of the fluid, and (c) the specific volume of the fluid. 2.12: A fluid mass of 13.5 kg completely fills a 12 L container. Given the acceleration of gravity as 9.81 m/s2 , determine (a) its density, (b) its specific volume, (c) its specific weight, and (d) its specific gravity if the density of water is 1000 kg/m3 . 2.13: A pump discharges 75 gal/min of water whose density is 998 kg/m3 . Find (a) the mass flow rate of the water in kg/s and (b) the time required to fill a cylindrical vat that is 6 m in diameter and 5 m high. 2.14: A fluid mass of 20 kg has a density of 2 kg/m3 . Determine (a) its volume, (b) its specific volume, (c) its specific weight, and (d) its weight. 2.15: The mass of the earth has been estimated as 5.981 × 1024 kg and its mean radius is 6.38 × 106 m. Estimate its apparent density. 2.16: Suppose that 2 m3 of liquid Awith a specific gravity of 1.24 and 3 m3 of liquid B with a density of 880 kg/m3 are completely mixed. Determine (a) the density of the mixture and (b) the weight of the mixture contained in a volume of 7.38 m3 .

Thermodynamics: Preliminary Concepts and Definitions

35

2.17: A 150-kg mass has a uniform density of 4250 kg/m3 . Determine (a) its weight, (b) its volume, and (c) its specific volume. 2.18: A liquid has a specific volume of 8 × 10−4 m3 /kg. Taking the acceleration of gravity as 9.80 m/s2 and the density of water as 1000 kg/m3 , determine (a) its density, (b) its specific weight, and (c) its specific gravity. 2.19: A glass tube with an inner diameter of 1 cm contains 8 g of water ( = 1000 kg/m3 ) and 14 g of gage oil with a specific gravity of 1.40. Taking the acceleration of gravity as 9.807 m/s2 , determine the height of the liquid in the tube. 2.20: A cylindrical container with a diameter of 2 m contains three fluids: Fluid Water Oil Mercury

, kg/m3

SG

, N/m3

Weight, N

1420

27,500 2000 40,000

1000 13.6

Determine the total height of the fluids in the cylinder. Pressure 2.21: A pipe with an inner diameter of 3 cm and a wall thickness of 2.286 mm contains air at a pressure of 27.6 kPa. Determine the tensile stress in the pipe wall. 2.22: A penstock, 1.75 m in diameter is built of longitudinal wood staves held together by circumferential steel hoops spaced 10 cm center-to-center apart. The water pressure is 56 kPa and the allowable tensile stress in the hoops is 130 MPa. The hoops are threaded for nuts. Determine their root diameter. 2.23: A wood stave pipe, 75 cm in diameter, is to operate under a water pressure of 350 kPa. The pipe will be wound spirally with 0.635 cm diameter steel wire at a tensile strength of 168 MPa. What spacing of wire is required? 2.24: A composite vertical fluid column that is open to the atmosphere is composed of 8 cm of mercury with a specific gravity of 13.6, 12 cm of water with a density of  = 1000 kg/m3 , and 14 cm of oil with a specific gravity of 0.831. Taking the acceleration of gravity as 9.81 m/s2 , determine (a) the pressure at the base of the column, (b) the pressure at the oil-water interface, and (c) the pressure at the water-mercury interface. 2.25: Can you determine the density of mercury from the observation that a pressure of 1.01325 bar supports a column of mercury that is 760 mm high? 2.26: Determine the absolute pressure exerted by a liquid having a depth of 4 m and a density of 875 kg/m3 when the atmospheric pressure is 1 bar and the acceleration of gravity is 9.804 m/s2 . 2.27: The pressure of a system is indicated by a pressure gage reading a vacuum pressure of 600 mbar when the atmospheric pressure is 1.025 bar. Determine the absolute pressure of the system. 2.28: Figure P2.28 shows a piston whose mass is 160 kg, which is raised on the inside of a 40 cm diameter vertical cylinder. The lower end of the cylinder is held in a pool of water whose density is 1000 kg/m3 . The pool of water is exposed to the atmosphere at 1.034 bar and the acceleration of gravity is 9.76 m/s2 . The water rises to a height of 6 m as shown and friction is to be neglected. Find (a) the pull on the piston and (b) the pressure exerted by the water on the piston.

36

Introduction to Thermal and Fluid Engineering Atmospheric Pressure

F P

40 cm 6m

FIGURE P2.28

2.29: A vertical tube contains 86 cm of mercury having a specific gravity of 13.6 and a layer of water at a density of 875 g/m3 . Taking the acceleration of gravity as 9.81 m/s2 , determine the height of the water layer if the tube is open to the atmosphere and the pressure at the bottom of the mercury layer is 1.225 bar. 2.30: Determine the height of a column of water at a density of 1000 kg/m3 that can be supported by the atmosphere at 1.0135 Pa with the acceleration of gravity equal to 9.807 m/s2 . Temperature 2.31: At what point are the Celsius and Fahrenheit temperatures identical? 2.32: Fahrenheit and Celsius thermometers are immersed in a fluid and give readings such that the Fahrenheit reading is 2.2 times the Celsius reading. Determine the two readings. 2.33: Convert a reading of 77◦ F to K. 2.34: Convert a reading of 86◦ C to ◦ R. 2.35: A new proposed absolute temperature scale is known as the Beaver scale (◦ B). In the Beaver scale, water turns to ice at 6279◦ B. Determine (a) the boiling point of water and (b) the temperature in ◦ C at 10, 810◦ B. 2.36: Define a new linear temperature scale, say ◦ Z, where the freezing and boiling points of water are 200◦ Z and 500◦ Z. Determine a correlation between the Z scale and the Celsius scale. 2.37: An element boils at 480◦ C. Using the Z scale developed in Problem 2.36, determine the boiling point in ◦ Z. 2.38: If a substance boils at 800◦ Z (Problem 2.36), determine the boiling point in ◦ C. 2.39: Determine the Celsius equivalent of 248◦ F. 2.40: Determine the Fahrenheit equivalent of 380◦ C.

3 Energy and the First Law of Thermodynamics

Chapter Objectives •

To describe the forms of energy and to define what is meant by kinetic, potential, internal, and total energy.

To define work and to show that it is not a property but a path function that depends on the path between two state points. • To define heat transfer and to show that it is not a property but a path function that depends on the path between two state points. • To present the concept of conservation of energy and to link all of the energy quantities considered into the first law of thermodynamics. • To consider thermodynamic cycles. •

•

To develop the ideal gas model.

•

To consider enthalpy and specific heats for ideal gases.

•

To present the equations that govern three of the five fundamental processes of the ideal gas.

3.1

Introduction

Energy may be defined as the capability or capacity to produce work. It is contained in all matter and while it exists in many different forms, these forms, however, are well defined. Because matter is anything that possesses mass and occupies space, energy is related to mass. Moreover, we may note that Einstein’s theory of relativity suggests that mass, m, may be converted to energy, E, (and energy may be converted to mass) via E = mc 2 where c = 2.9997 × 108 m/s is the speed of light. However, for all energy-mass interactions other than nuclear reactions, the amount of mass converted to energy is extremely small and can be neglected. Thus, in this study, we state the conservation of mass principle that is often quoted in subsequent discussions as Mass can neither be created nor destroyed and its composition cannot be altered from one form to another unless it undergoes a chemical change. 37

38

Introduction to Thermal and Fluid Engineering

FORMS OF ENERGY Kinetic Energy is the energy that a body possesses by virtue of its motion. Potential Energy is the energy that a body possesses by virtue of its position. Internal Energy is a characteristic property of the state of a thermodynamic system including the intrinsic energies of individual molecules, kinetic energies of internal motions, and contributions from interations between individual molecules.

FIGURE 3.1 Forms of energy. The internal energy is treated as a macroscopic quantity even though it is composed of many microscopic energies resulting from the atomic and molecular structure.

Having thus observed that mass is a conserved property, we next focus our attention on the forms of energy and a general conservation of energy principle known as the first law of thermodynamics. The forms of energy are summarized in Figure 3.1.

3.2

Kinetic, Potential, and Internal Energy

Macroscopic kinetic energy, gravitational potential energy (most often referred to merely as “potential energy”), and microscopic internal energy are scalar properties and we examine them in some detail in the subsections that follow. 3.2.1 Kinetic Energy In Newton’s second law, F = ma , the acceleration can be represented by a=

d Vˆ d Vˆ dx d Vˆ = = Vˆ dt d x dt dx

where x is the displacement and Vˆ is the velocity in the direction of the applied force, F. If this form of the acceleration is substituted into F = ma where the applied force is constant, the result is d Vˆ F = m Vˆ dx or, after separation of the variables ˆ Vˆ F d x = m Vd Integration between x1 , where the velocity is Vˆ 1 , and x2 , where the velocity is Vˆ 2 , gives Fs =

m ˆ2 2 ( V − Vˆ 1 ) 2 2

Energy and the First Law of Thermodynamics

39

Vˆ

z

Datum Level FIGURE 3.2 A projectile that possesses kinetic energy by virtue of its velocity and potential energy by virtue of its elevation, z, above the datum level. 2 where s = x2 − x1 . The product of m and Vˆ /2 leads to the kinetic energy (Figure 3.2) defined as

m Vˆ KE = 2

2

(N-m or joules)

(3.1)

and the kinetic energy may also be written as a specific intensive property 2 KE Vˆ ke = = m 2

(N-m/kg or joules/kg)

(3.2)

3.2.2 Potential Energy Potential energy is energy that is associated with a system by virtue of its position in a force field and gravitational potential energy is associated with the gravitational force field. Potential energy is stored in a system because of its elevation above some arbitrary datum (Figure 3.3) and is potentially available for conversion to work. The change in potential energy, d(PE), is equal to d(PE) = Wdz = mgdz Top of Arc Minimum KE Maximum PE

Datum Level

FIGURE 3.3 A projectile in a semicircular arc.

Impact Point Maximum KE PE = 0

40

Introduction to Thermal and Fluid Engineering

where W is the weight of the system and z is its elevation above the datum. An integration in a constant force field gives (PE) = mg(z2 − z1 ) where z1 and z2 represent the initial and final elevations. Hence, potential energy is given by PE = mgz

(N-m or joules)

(3.3)

Potential energy is also a scalar property and may be written as a specific intensive property pe =

PE = gz m

(N-m/kg or joules/kg)

(3.4)

We note that potential energy does not have an absolute value and that only changes in potential energy with respect to some arbitrary datum are of interest. 3.2.3 Internal Energy The internal energy of a substance is energy that is associated with the motion and configuration of its molecules, atoms, and subatomic particles such as nuclei and protons. It is a property consisting of the combined molecular kinetic and potential energies due to the random motion, vibration, rotation, and the intermolecular forces between the molecules and can be determined from other properties such as temperature and pressure. The internal energy is designated by the letter U and because it is a unit of energy, it is expressed in N-m or joules. Specific internal energy, which is an intensive property, is u=

U m

(N-m/kg or joules/kg)

(3.5)

3.2.4 The Total Energy The total energy possessed by a substance is the sum of the kinetic, potential, and internal energies E = KE + PE + U

(3.6)

The total energy is a system property, and we illustrate this idea in Example 3.1.

Example 3.1 The datum for a moving mass of 50 kg (Figure 3.4) is taken at sea level

(the surface of the earth) where the acceleration of gravity is 9.81 m/s2 . The mass is at an elevation of 175 m and has a total energy of 2068 kJ with a specific internal energy of u = 28.44 kJ/kg. Determine the velocity of the mass.

Solution Assumptions and Specifications (1) The 50-kg mass forms the system. (2) The system is moving with uniform velocity. (3) The local acceleration of gravity is 9.81 m/s2 . (4) The elevation, total energy, and specific internal energy are all specified.

Energy and the First Law of Thermodynamics

41

Total Energy E = 2068 kJ Specific Internal Energy u = 28.44 kJ/kg ˆ V

50 kg

175 m

Datum Level FIGURE 3.4 The moving mass in Example 3.1.

The strategy here is to compute the potential energy and then use the total and internal energies to find the kinetic energy. We employ a modification of Equation 3.6 KE = E − PE − mu and with PE = mgz = (50 kg)(9.81 m/s2 )(175 m) = 85.84 × 103 kg-m2 /s2 = 85.84 kJ Then with E = 2068 kJ, we have KE = 2068 kJ − 85.84 kJ − (50 kg)(28.44 kJ/kg) = 2068 kJ − 85.84 kJ − 1422 kJ = 560.16 kJ The velocity can be obtained from this value of kinetic energy 2

KE =

1 ˆ2 (50 kg) Vˆ mV = = 560.16 kJ 2 2(1 kg-m/N-s2 )

Thus, 2(560.16 × 103 N-m)(1 kg-m/N-s2 ) 2 Vˆ = = 22, 406 m2 /s2 50 kg and Vˆ = 149.7 m/s ⇐

42

3.3

Introduction to Thermal and Fluid Engineering

Work

Work is defined as the energy that is expended when a force acts through a displacement  W=

 F · dx =

F cos  d x

(3.7)

where  is the angle between the line of action of the force and the direction of the displacement (Figure 3.5). We note that in F · dx, F, and dx are vectors and, because the scalar or dot product yields a scalar, work is a scalar quantity. Equation 3.7 defines work in the macroscopic mechanical terms of force and displacement and, because the force may not be constant, we see that the indicated integration may require a knowledge of how the force varies with the displacement. We note that the thermodynamic view of work is that it is a form of energy transfer between a system and its surroundings. Thus, it is identified at the system boundary where its sole external effect is equivalent to the raising or lowering of a weight. Of course, in the real world, the raising or lowering of a weight may never be observed and the test is really whether or not the sole effort on the surroundings “could be” the raising or lowering of a weight. The foregoing considerations for work show that it is energy that crosses a system boundary. Thus, we see that it is a transfer phenomenon and we will designate work as being positive if it is done by the system and negative if it is done on the system  Work −→

done by a system is positive done on a system is negative

Consider Figure 3.6a, which shows a compressible gas within a piston-cylinder assembly. The gas at state 1 is compressed to state 2 and the path is shown in pressure and volume coordinates that we refer to as the P-V plane in Figure 3.6b. With the differential amount of work designated as W, W = −Fdx where the force on the piston, F = PA, is in the opposite direction of the displacement, dx. Hence, W = −PA dx and because the differential volume is dV = Adx, we see that W = −PdV F

θ M

dx FIGURE 3.5 A vector force F on a moving mass, m, such that the displacement to the right is dx.

Energy and the First Law of Thermodynamics 2

43

1

F

dV = Adx (a) P P2, V2

δW = PdV

P1, V1 V (b)

FIGURE 3.6 Work in a quasi-equilibrium compression process: (a) the piston-cylinder arrangement and (b) the P-V diagram.

and  W=−

V2

PdV

(3.8)

V1

If a functional relationship between P and V is known, Equation 3.8 can be employed to evaluate the work in this quasi-equilibrium process. Reference to Figure 3.6b shows W as the shaded strip and the integral indicates that the entire area under the curve between V1 and V2 is equal to the work done on the system by the surroundings. We note that had the process proceeded in the opposite direction, the magnitude of the work would have remained the same but the work would have been positive indicating that work was done by the gas on the surroundings. Indeed, a different functional relationship between P and V would have led to a different path between the two end states. We are led by the foregoing development to the conclusion that work depends on the process between the end states of a system and that systems may possess energy but they cannot possess work. Thus, work at a system boundary is not a system property and different P-V relationships will yield different amounts of work. Because the magnitude of the work depends on the path between the initial and final states, work is considered to be a path function, which has an inexact differential that we have designated as W.

Example 3.2 The initial and final volumes of a gas are 8 m3 and 16 m3 . If the initial pressure is 20 N/m2 , determine the work done if (a) the pressure during expansion is constant and (b) the pressure during the expansion process is inversely proportional to the volume, that is, P V = C (a constant).

44

Introduction to Thermal and Fluid Engineering

Solution Assumptions and Specifications (1) The gas in the enclosure is taken as a closed system. (2) Kinetic and potential energy changes are negligible. (3) The displacement of the moving boundary represents the only work mode. (4) The expansion is governed by a series of quasi-static equilibrium states. (a) With P constant  W=

V2

 Pd V = P

V1

V2

dV V1

16 m3  = (20 N/m )V  8 m3 2

= (20 N/m2 )(16 m3 − 8 m3 ) = 160 N-m

(160 J) ⇐

(b) With P = C/V or P V = C  W=

V2

 Pd V = C

V1

V2 V1

dV V

V2  16 m3 = C ln V  = C ln 8 m3 V1

= C ln 2 However, P V = P1 V1 = C so that C = (20 N/m2 )(8 m3 ) = 160 N-m and with ln 2 = 0.6931 W = 0.6931(160 N-m) = 110.9 N-m

(110.9 J) ⇐

In both cases (a) and (b), an expansion of the gas is indicated (Figure 3.7). Under this circumstance, the work is done by the system and is positive. Figure 3.7 illustrates what the example shows in a straightforward manner; that the work done by the system varies as the functional relationship between the P and V values. Moreover, work is a path function and not a system property. The work in an expansion or compression quasi-equilibrium process is but one example of a work interaction between a system and its surroundings. Other examples include electrical and magnetic work, extension, and compression of a spring, work in stretching a film or membrane and work transmitted by a rotating shaft. We will discuss shaft or paddle wheel work, electrical work, and the work involved in making a bar elongate and the stretching of a spring in Section 3.6.2.

Energy and the First Law of Thermodynamics

45

P

2 Path-a -b th Pa

Path-c 1

V1

V2

V

FIGURE 3.7 Work is a path function and not a system property. These three paths enclose different areas in the P-V plane and all yield different amounts of work.

3.3.1 Power Power is the rate of doing work ˙ = dW W dt

(3.9)

and the unit of power is the watt (W) 1 W = 1 J/s

or

1 N-m/s

Power in a translational system can be expressed as ˙ trans = dW = d(Fx) = F dx = F Vˆ W dt dt dt

(watts)

(3.10a)

and in a rotational system where the work is the product of the torque, , in joules and the angular displacement, , in radians ˙ rot = dW = d() =  d =  W dt dt dt

3.4

(watts)

(3.10b)

Heat

All energy transfer, other than work, across the boundaries of a system is accounted for by heat,1 Q. This heat is also a form of energy and its flow is induced by a temperature difference between the system and the surroundings and it will flow in the direction of

1 There

is a huge difference between heat and heat transfer. Here, we consider heat as energy, and in Chapter 5, ˙ where we deal with open systems, we consider the rate of heat transfer, Q.

46

Introduction to Thermal and Fluid Engineering

decreasing temperature. The transfer of heat is taken as positive when it is transferred to the system and negative when it is transferred from the system  Heat −→

transferred to a system is positive transferred from a system is negative

From this, we note that the sign convection pertaining to heat transfer is exactly opposite to the sign convention adopted for the work. Like the work interaction, the heat interaction depends on the path between the end ˙ states of a process and is not a property. The rate of heat transfer, already designated as Q in watts or J/s, is related to the heat energy, Q, by ˙ = Q Q dt

watts

(3.11)

where Q, like W is an inexact differential. The heat flux, q˙ , is the rate of heat transfer per ˙ by unit surface area on the system boundary and, just as Q is related to Q  ˙ dt Q= Q (joules) ˙ is related to q˙ by Q ˙ = Q

 q˙ d S

(watts)

where S is the surface area on the system boundary through which the heat passes.

3.5

The First Law of Thermodynamics

We now consider a process in which both heat and work interactions (Q and W) occur between a system and its surroundings. For the closed system or control mass shown in Figure 3.8, we modify Equation 3.6 to include both heat and work, as follows: E2 − E1 = Q − W

(3.12)

This is the first law of thermodynamics that can be written in a differential form d E = Q − W

(3.13)

Both Equations 3.12 and 3.13 indicate that energy within a closed system can be changed only by means of a transfer of energy by heat or work between the system and its surroundings. As a word statement, the first law of thermodynamics is

Q

System E2 – E1

W

FIGURE 3.8 A system with both heat and work interaction between the system and its surroundings.

Energy and the First Law of Thermodynamics

47

The change of total energy (kinetic, potential, and internal) is equal to the net heat transferred to the control mass minus the work done by the control mass. We may also observe that either of Equations 3.12 and 3.13 demonstrate the conservation of energy principle: Energy may neither be created nor destroyed but may be converted from one form to another.

3.6

The Energy Balance for Closed Systems

3.6.1 Processes Equation 3.12 E2 − E1 = Q − W

(3.12)

indicates that over a time interval ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎪ the net amount of ⎪ ⎪ the net amount of ⎪ ⎪ ⎪ ⎪ ⎨ the change in ⎬ ⎨ ⎬ ⎪ ⎨ ⎬ energy transferred energy transferred energy content = − into the system ⎪ ⎩ ⎭ ⎪ ⎪ out of the system ⎪ ⎪ ⎪ ⎪ within a system ⎩ ⎭ ⎪ ⎩ ⎭ boundary as heat boundary as work

(3.13)

Equation (3.12) may be written using total or specific quantities. We will find it convienient to use a form that places the properties at state 1 and the heat transfer on the left-hand side and the properties at state 2 and the work interaction on the right-hand side. This is consistent with the conventions that we have adopted for the direction of the heat flow and whether work is done on or by the system. Thus, in terms of total energy, we have 1 ˆ2 1 2 m V 1 + mgz1 + U1 + Q = m Vˆ 2 + mgz2 + U2 + W 2 2

(J)

(3.14a)

or in terms of specific energy, 1 ˆ2 Q 1 2 W V 1 + gz1 + u1 + = Vˆ 2 + gz2 + u2 + 2 m 2 m

(J/kg)

(3.14b)

We may write Equation 3.13 as E = Q − W Then, over some time interval, this becomes E Q W = − t t t and in the limit as t −→ 0 dE ˙ −W ˙ =Q dt With dE dKE dPE dU = + + dt dt dt dt

(3.15)

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Introduction to Thermal and Fluid Engineering

we see that Equation 3.15 can be written as dKE dPE dU ˙ −W ˙ + + =Q dt dt dt

(3.16)

Example 3.3 During the execution of a certain process, the work done per degree of temperature increase is 60 J/K and the internal energy is given as a function of temperature, U = 25(1 + 0.16T) J/kg. Determine the heat transferred in joules if the temperature changes from 325 K to 400 K.

Solution Assumptions and Specifications (1) We consider the working medium as the closed system. (2) Kinetic and potential energy changes are negligible. (3) The process is represented by a series of quasi-equilibrium processes. (4) The internal energy is a specified function of the temperature. Here, from Equation 3.14a we have U1 + Q = U2 + W and because W and U = U2 − U1 can be evaluated Q = W + U = W + U2 − U1 With dW/dT = 60 J/K  W=

T2

 (60 J/K)dT = (60 J/K)

T1

400 K 325 K

dT = (60 J/K)(75 K) = 4500 J

Moreover, U = 25(1 + 0.16T2 ) J/K − 25(1 + 0.16T1 ) J/K = (4.0 J/K)(400 K − 325K) = (4.0 J/K)(75 K) = 300 J Hence, Q = W + U2 − U1 = 4500 J + 300 J = 4800 J ⇐ and this positive value indicates that this heat is transferred to the system.

Example 3.4 An elevator weighs 4750 N and is moving downward at a velocity of 1.92 m/s (Figure 3.9). The elevator has a 3250-N counterweight and an associated braking system. Determine the frictional energy absorbed by the braking system when the elevator is brought to a stop in a distance of 2 m.

Energy and the First Law of Thermodynamics

49

ˆ = 1.92 m/s V Elevator Counterweight

FIGURE 3.9 Elevator, counterweight, and rotating pulley for Example 3.4.

Solution Assumptions and Specifications (1) Both the elevator and the counterweight are taken as the system. (2) Rotating kinetic and potential energies in all cables and pulleys in the system are negligible. (3) The local acceleration of gravity is taken as g = 9.81 m/s2 . With the subscripts E, CW, and B designating the elevator, counterweight, and braking system respectively, conservation of energy dictates that KEE + PEE = KECW + PECW + WB With Vˆ = 1.92 m/s for both the elevator and the counterweight

1 1 4750 N 2 KEE = m Vˆ = (1.92 m/s) 2 = 892.5 J 2 2 9.81 m/s2 and for a change in elevation of 2 m

4750 N PEE = mgz = (9.81 m/s2 )(2 m) = 9500 J 9.81 m/s2 For the counterweight by straight proportion

3250 N KECW = (892.5 J) = 610.7 J 4750 N and

PECW =

3250 N (9500 J) = 6500 J 4750 N

We may now solve for the braking energy WB = KEE − KECW + PEE − PECW = 892.5 J − 610.7 J + 9500 J − 6500 J = 3281.8 J ⇐

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Introduction to Thermal and Fluid Engineering

FIGURE 3.10 Rotating paddle wheel and shaft inside an enclosure.

3.6.2 Other Forms of Work 3.6.2.1 Shaft or Paddle Wheel Work Consider Figure 3.10, which shows a closed, rigid, and insulated container holding a mass of gas, m, at a temperature, T1 . A shaft passes through the container wall and the shaft is rotated to spin the paddle wheel (or vane or rotor). We may write the first law energy balance E2 − E1 = Q − W

(3.12)

and we note that because the container is closed, rigid, and insulated, Q = W = 0. Under these circumstances E 1 = E 2 or U1 = U2 . Because, as indicated in Section 3.2.3, the internal energy is a function of temperature, this further infers that T = 0. A little thought leads us to the conclusion that energy is delivered to the mass of gas inside of the container by the rotation of the shaft. This energy is called shaft or paddle wheel work Wp =  where  is the average torque applied to the shaft and  is the angular displacement. Clearly, the first law must be generalized to accommodate other kinds of work in addition to the boundary work (the energy that flows into or out of the system across the system boundaries). Moreover, the power due to the paddle wheel will be ˙ p =  W where  is the angular velocity of the shaft. 3.6.2.2 Electrical Work A study of electrostatics will show that work is done on or by an electric charge as it is moved through a potential difference V. Thus, We = Vq = (V2 − V1 )q and if V1 = 0, we have with V2 = V We = Vq

Energy and the First Law of Thermodynamics

51

I + V –

R

FIGURE 3.11 Current flow in a resistor.

Because instantaneous current, in amperes, is the rate of flow of charge i≡

dq dt

and with v taken as the instaneous voltage we have the instantaneous electric power ˙ e = v dq = vi W dt If the system is time invariant (Figure 3.11), the electrical work over a time interval, t, will be ˙ e = VI W 3.6.2.3 Spring Work A spring with a spring constant, ks , in N/m can be stretched (Figure 3.12a) from an original length, x0 , to x = x1 − x0 , so that the tension force will be Ft = −ks (x1 − x0 ) where ks > 0. If the spring is compressed (Figure 3.12b) from its original length, x0 , to x = x2 + x0 , the compression force is Fc = ks (x2 − x0 )

x0

x2

x1

x0

F (a)

F (b)

FIGURE 3.12 A spring (a) stretched and (b) compressed from an original length, xo .

52

Introduction to Thermal and Fluid Engineering Clamp Bar

F

FIGURE 3.13 A solid bar under tension.

Here x = x1 − x0 − (x2 − x0 ) = x1 − x2 and the work done on the spring in changing its length from x1 to x2 will then be  W=

x2

 F x d x = ks

x1

x2

x dx =

x1

ks 2 (x − x12 ) 2 2

(W)

3.6.2.4 Two Additional Examples Two other examples of the work concept are the extension of a solid bar and the stretching of a liquid film. 3.6.2.4.1 The Stretching of a Solid Bar In Figure 3.13, one end of the solid bar is under tension and the other end is fixed at x = xo . When a force is applied at a point where the unstressed length is x1 , the bar will elongate to a length, x2 . With n taken as the normal stress in the bar and A the cross-sectional area, the work done will be  x2  x2 W=− F dx = − n Ad x = (x1 − x2 )n A x1

x1

where the minus sign is needed because work is done on the bar when d x is positive. 3.6.2.4.2 The Stretching of a Liquid Film A liquid film suspended in a rigid wire frame is shown in Figure 3.14. Here, the microscopic forces between the molecules near the liquidair interfaces support the thin liquid layer within the frame by the effect of surface tension. These microscopic forces establish a measurable macroscopic force that is perpendicular to any line in the liquid film and the force per unit length in such a line is the surface tension, . Thus, the force, F, shown in Figure 3.14 can be written as F = 2L because there are two film surfaces that act on the wire. If the movable wire is displaced a distance, x, the work is given by  W=−

x2

F d x = 2L(x1 − x2 )

x1

where the minus sign is required because the work is done on the system when d x is positive.

Energy and the First Law of Thermodynamics

53

Moveable Member

F

L

Surface of Film FIGURE 3.14 A liquid film suspended in a rigid wire frame.

3.6.3 Cycles The study of devices that operate on a cycle is an important application of thermodynamics. Because the working substance returns to its initial state, E = 0 and Equation 3.12 may be written as Q−W =0 or with emphasis on the application to a cycle Qcycle = Wcycle

(3.17)

Figure 3.15 shows two types of cycles and the directions of all heat and work interactions are indicated by arrows. We note that the heat interactions derive from an external body at a “hot side” temperature, TH , and an external body at a “cold side” temperature, TL . For the power cycle shown in Figure 3.15a, the net work is Wcycle = Qin − Qout and we observe that in order to have work done by the system in a power cycle, Qin must exceed Qout . In this case, the cycle efficiency is defined as the ratio of the work done by the cycle to the heat input =

Wcycle Qin − Qout Qout = =1− Qin Qin Qin

(3.18)

and is less than unity. Notice that, because a cycle begins and ends at the same state point, when dealing with cycles where only Wcycle , Qout , and Qin are involved, each of these quantities can be represented by a positive number. In this event, any negative signs in Equation 3.17 will automatically take care of any subtraction. The use of positive numbers to represent Qout , and Qin will be followed in subsequent chapters when cycles (with E 1 = E 2 ) are considered.

54

Introduction to Thermal and Fluid Engineering TH Qin

System

W = Qin – Qout

Qout TL (a)

TH Qout

W = Qout – Qin Qin TL (b) FIGURE 3.15 Sketches of two cycles: (a) power cycle and (b) refrigeration or heat pump cycle.

The cycle in Figure 3.15b is either a refrigeration or a heat pump cycle. If it is to represent a refrigerator, its purpose is to remove heat from a body or system (such as a refrigerated space) at TL . If it is to represent a heat pump, its purpose is to provide heat to a body or system (such as a dwelling) at TH . In the case of either the refrigerator or the heat pump Wcycle = Qout − Qin and Qout must exceed Qin . In both, the coefficient of performance is a measure of the effectiveness of the device or system. For the refrigerator =

Qin Qin = Wcycle Qout − Qin

(3.19)

=

Qout Qout = Wcycle Qout − Qin

(3.20)

and for the heat pump

Example 3.5 A thermal cycle contains an engine that delivers 24 horsepower (HP) to a load and uses a small pump, which absorbs 1.85 KW to circulate the fluid in the system (Figure 3.16). If the heat supplied in a heat exchanger is 90 KW, determine (a) the net work, (b) the heat rejected, and (c) the efficiency of the cycle.

Solution Assumptions and Specifications (1) The working medium is considered as the system. (2) The system operates in the steady state.

Energy and the First Law of Thermodynamics

55

Engine 24 HP Heat Exchanger

1.85 kW

90 kW

FIGURE 3.16 Thermal cycle for Example 3.5.

(a) The engine delivers Wout = (24 HP)(0.746 kW/HP) = 17.90 kW and the net work, which will be Wcycle , is Wcycle = Wout − Wpump = 17.90 kW − 1.85 kW = 16.05 kW ⇐ (b) The heat rejection is obtained from Wcycle = Qin − Qout so that Qout = Qin − Wcycle = 90 kW − 16.05 kW = 73.95 kW ⇐ (c) The efficiency is =

3.7

Wcycle 16.05 kW = = 0.178 Qin 90 kW

or

17.8 % ⇐

The Ideal Gas Model

From our study of physics, we may recall Charles law,2 which derived from a consideration of the relationships between pressure and volume for a gas. These relationships are idealizations and give very close approximations to the behavior of real gases at low pressures and moderate temperatures. We may consider the ideal gas model by referring to the alternate forms of Charles’ law: At constant pressure, the volume of a fixed mass of a gas varies directly with its absolute temperature.

2 J. A. C.

Charles (1746–1823) was a French chemist, physicist, and inventor.

56

Introduction to Thermal and Fluid Engineering P P=C

1

2

v=C

3 v FIGURE 3.17 The P-v plane with a constant pressure process followed by a constant volume process.

and At constant volume, the pressure of a fixed mass of a gas varies directly with its absolute temperature. Consider Figure 3.17, which shows the two alternative forms of Charles’ law in the P-v (pressure-specific volume) plane. We may note that the process from state 1 to state 2 is at constant pressure and that the process from state 2 to state 3 is at constant specific volume. From state 1 to state 2, Charles’ law states that v1 T1 = v2 T2

so that

T2 = T1

v2 v1

P2 T2 = P3 T3

so that

T2 = T3

P2 P3

and from state 2 to state 3,

Hence, T1

v2 P2 = T3 v1 P3

T1

v3 P1 = T3 v1 P3

and because P1 = P2 and v2 = v3

or P1 v1 P3 v3 PVC = = T1 T3 T

(3.21)

Because P, v, and T are all independent variables, each side of Equation 3.21 must be uniquely equal to a constant, R (called the gas constant), so that P V = RT

(3.22)

P V = mRT

(3.23)

and with v = V/m

Energy and the First Law of Thermodynamics

57 N2 Given M, P, T

Ideal Gas Law

Look up R

Obtain V FIGURE 3.18 The procedure for finding the volume of nitrogen in a closed, rigid container in Example 3.6.

Equations 3.22 and 3.23 are alternative expressions of the equation of state for the ideal gas, which is also referred to as the ideal gas law. In order to make this equation of state universal, the gas constant is related to another gas constant that holds for all gases via R=

R¯ M

(3.24)

where M is the molecular weight of the gas. Here, R¯ is called the universal gas constant. Because the mass is related to the number of moles, n, by the molecular weight n=

m M

we can provide another representation for the equation of state PVC = m

R¯ ¯ T = n RT M

(3.25)

Moreover, because the density of the gas is = m/V, the equation of state can also be written as P = RT

(3.26)

¯ is based on Avogadro’s law3 which states that The value of the universal gas constant, R, one mole of a gas at standard conditions of temperature and pressure contains 6.023 × 1023 molecules R¯ = 8314 Pa-m3 /kgmol-K

or

R¯ = 8314 J/kgmol-K

(3.27)

Values of the gas constant, R, for several gases are listed in Table A.1 in Appendix A, which also lists values of the molecular weight and specific heats that are considered in the next section. 3 Amadeo

Avogadro ( 1776–1856) was an Italian physicist.

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Introduction to Thermal and Fluid Engineering

Example 3.6 32 kg of nitrogen at 1.2 MPa and 177◦ C are in a closed, rigid container. Use the procedure provided in Figure 3.18 to determine the volume of the container.

Solution Assumptions and Specifications (1) We assume that the nitrogen is the system. (2) The nitrogen is an ideal gas. (3) Kinetic and potential energy changes are negligible. From Table A.1, we read R = 297 J/kg-K. Then with T = 117◦ C + 273◦ C = 390 K, the volume of the container will be V= =

mRT P (32 kg)(297 N-m/kg-K)(390 K) 1.2 × 106 N/m2

= 3.089 m3 ⇐

3.8

Ideal Gas Enthalpy and Specific Heats

Because the internal energy, U, and the product, P V, occur so frequently in so many thermodynamic considerations, their sum is given the name enthalpy, designated by H H = U + PV

(J)

(3.28a)

(J/kg)

(3.28b)

or specific enthalpy designated by h h=

H = u + Pv m

The specific heat of a substance may be defined as The quantity of energy that is required to raise a unit mass of the substance by one degree. For a closed system containing a gas and with no kinetic or potential energy change (a stationary system), Equation 3.12 tells us that in a constant volume process where no work is done by or on the gas Q = dU Here, Q is the energy transferred into the system so that we may define the specific heat at constant volume on a unit mass basis as  du  c v (T) = (J/kg-K) (3.29a) dT v = constant and we note that the specific heat, c v , is a function of temperature.

Energy and the First Law of Thermodynamics

59

Similar reasoning leads to the constant pressure process where we find that from Equation 3.28a d H = dU + d( P V) = dU + W and from Equation 3.12 Q = d H so that the specific heat at constant pressure, c p , also defined on a unit mass basis is  dh  c p (T) = (J/kg-K) (3.29b) dT  P = constant which is also a function of temperature. Now we know that u is a function of temperature (Section 3.2.3) and we see from Equation 3.28b that, because R is a constant, h = u(T) + RT = h(T) From Equation 3.29a, we may write du = c v (T)dT

(3.30a)

dh = c p (T)dT

(3.30b)

and from Equation 3.29b

Equations 3.30 may be integrated yielding  u2 − u1 =

T2

c v (T)dT

(3.31a)

c p (T)dT

(3.31b)

T1

and

 h2 − h1 =

T2 T1

If the specific heats are constant, the result of the integrations in Equation 3.31 will be u2 − u1 = c v (T2 − T1 )

(3.31c)

h 2 − h 1 = c p (T2 − T1 )

(3.31d)

and

Specific heats at constant volume and constant pressure for ideal gases may be taken as constant, determined from a table of specific heat values such as Table A.2 or evaluated from an equation such as the polynomial equation for the molar constant pressure specific heat c¯ p = a + bT + cT 2 + dT 3 + eT 4 R¯ We also note that the ideal gas law permits the representation h = u + Pv

−→

h = u + RT

60

Introduction to Thermal and Fluid Engineering

and because dh du = +R dT dT we see that c p (T) = c v (T) + R

(3.32)

R = c p (T) − c v (T)

(3.33)

or

Equation 3.33 shows that, because R is greater than 0, c p (T) must exceed c v (T). In subsequent sections, we will encounter the ratio of specific heats, c p /c v , designated as k. Some useful relationships are k= then

 R = c v (T)

c p (T) >1 c v (T)

(3.34)

 c p (T) − 1 = c v (T)(k − 1) c v (T)

so that c v (T) =

R k−1

(3.35a)

c p (T) =

kR k−1

(3.35b)

and

Table A.2 provides specific heats and their ratio, k, as a function of temperature for air and five other gases.

Example 3.7 Air at one atmosphere (101.3 kPa) and 25◦ C is contained in a piston-cylinder

assembly (Figure 3.19) where the volume is initially at 1.60 × 10−3 m3 . Heat is added in a manner such that the movement of the piston holds the air at constant pressure until the air reaches 275◦ C. Determine (a) the work done by the gas, (b) the heat added to the gas, and (c) the change in internal energy.

Q

P = 101.3 kPa T = 293 K V = 0.0016 m3

FIGURE 3.19 Air inside of a piston-cylinder assembly for Example 3.7.

AIR

W

Energy and the First Law of Thermodynamics

61

Solution Assumptions and Specifications (1) The air contained in the piston-cylinder assembly represents a closed system. (2) The expansion process is quasi-static. (3) Kinetic and potential energy changes are negligible. (4) The air may be considered as an ideal gas. (a) The work in a constant pressure process can be evaluated using Equation 3.8  W=

V2

Pd V = P(V2 − V1 )

V1

and because mR = P V/T, the volume at state 2 can be obtained from conditions at state 1 mR =

P1 V1 P2 V2 = T1 T2

and with P1 = P2 V2 =

T2 V1 T1

Here, T1 = 25◦ C + 273◦ C = 298 K, T2 = 275◦ C + 273◦ C = 548 K and the average temperature is Tavg = Then

V2 =

298 K + 548 K = 423 K 2

548 K (1.6 × 10−3 m3 ) = 2.942 × 10−3 m3 298 K

and W = P(V2 − V1 ) = (1.013 × 105 N/m2 )(2.942 × 10−3 m3 − 1.600 × 10−3 m3 ) = (1.013 × 105 N/m2 )(1.342 × 10−3 m3 ) = 136.0 J ⇐ Because the work is a positive quantity, we conclude that the work is done by the system (the gas in the cylinder) on the surroundings. (b) Because kinetic and potential energy changes are negligible, we can rearrange Equation 3.12 to show that Q = m(u2 − u1 ) + W For this constant pressure process Q = m(u2 − u1 ) + mP(v2 − v1 ) = m(h 2 − h 1 )

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Introduction to Thermal and Fluid Engineering

and we can compute the heat added to the gas from the change in enthalpy using Equation 3.31d h 2 − h 1 = c p (T2 − T1 )

(3.31d)

First, however, we will need the amount of air present. We read from Table A.1 for air at 298 K ≈ 300 K that R = 287 J/kg-K and we calculate m from conditions at point 1 using the ideal gas law. m=

PV (1.013 × 105 N/m2 )(1.60 × 10−3 m3 ) = = 1.895 × 10−3 kg RT (287J/kg-K)(298 K)

Then with T2 − T1 = 548 K − 298 K = 250 K and with Tavg = 423 K, Table A.2 reveals that c p = 1.016 kJ/kg-K and c v = 0.729 kJ/kg-K so that Q = m(h 2 − h 1 ) = mc p (T2 − T1 ) = (1.895 × 10−3 kg)(1016 J/kg-K)(250 K) = 481.4 J ⇐ (c) The change in internal energy can be computed from the energy equation of Equation 3.12 or from Equation 3.31a. First, from the energy equation U2 − U1 = Q − W = 481.3 J − 136.0 J = 345.3 J ⇐ and alternatively from Equation 3.31c with c v = 0.729 kJ/kg-K from Table A.2 at 423 K U2 − U1 = mc v (T2 − T1 ) = (1.895 × 10−3 kg)(729 J/kg-K)(250 K) = 345.3 J

3.9

Processes of an Ideal Gas

3.9.1 Introduction We will find it extremely useful to characterize quasi-static processes of an ideal gas by the relationship P Vn = C

(3.36)

where the exponent, n, may take on different values that are dictated by the particular process. We consider five such processes of the ideal gas, three of which are: •

The isometric or constant volume process where n = ∞

Energy and the First Law of Thermodynamics

63

The isobaric or constant pressure process where n = 0 • The isothermal or constant temperature process where n = 1 •

The other two, which are discussed in subsequent chapters: •

The isentropic or constant entropy process where n is equal to k = c p /c v

•

The polytropic process where n is unspecified and can take on any value between 0 and ∞

At this point in our study, we are able to treat the first three of these ideal gas processes with modest detail. 3.9.2 The Constant Volume Process The constant volume process is characterized by n = ∞ so that Equation 3.36 takes the form V=C and the pressure-temperature relationship is T2 P2 = T1 P1

(3.37)

Because V2 = V1 , Equation 3.8 is evaluated as  W=

V2

Pd V = 0

V1

which tells us that no work is done on or by the system in a constant volume process. Thus, in the absence of kinetic and potential energy changes, Equation 3.12 becomes U2 − U1 = Q and via Equation 3.31a  Q = U1 − U1 = m

T2

c v (T)dT

(3.38a)

T1

and for small temperature excursions where c v (T) can be taken as constant Q = mc v (T2 − T1 )

(3.38b)

3.9.3 The Constant Pressure Process The constant pressure process is characterized by n = 0 so that Equation 3.36 takes the form P =C and the volume-temperature relationship is T2 V2 = T1 V1

(3.39)

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Introduction to Thermal and Fluid Engineering

Here, by Equation 3.8, the work is 

V2

W=

 Pd V = P

V1

V2

d V = P(V2 − V1 )

(3.40)

V1

and if V2 > V1 , the work is done by the system on the surroundings. Finally, by Equation 3.12, we see that U2 − U1 = Q − W = Q − P(V2 − V1 ) Q = U2 − U1 + ( P V2 − P V1 ) = H2 − H1 or via Equation 3.31b  Q = H2 − H1 = m

T2

c p (T)dT

(3.41a)

T1

and for small temperature excursions where c p (T) can be taken as constant Q = H2 − H1 = mc p (T2 − T1 )

(3.41b)

3.9.4 The Isothermal Process The constant temperature process or isothermal process is characterized by n = 1 so that Equation 3.36 takes the form PV = C and the pressure-volume relationship is P1 V2 = P2 V1

(3.42)

Here, by Equation 3.8, the work is 

V2



V2

Pd V =

V1

V1

C V2 d V = C ln V V1

But C = P1 V1 = P2 V2 so that we have W = P1 V1 ln

V2 V2 = P2 V2 ln V1 V1

(3.43)

and we note that if V2 > V1 work is done by the system on the surroundings. Moreover, we note that because T2 = T1 Q = U2 − U1 = 0 and Q=W Thus, we see that in an isothermal process, work is done on the surroundings and must equal the heat transferred to the system because there is no change in internal energy in the system.

Energy and the First Law of Thermodynamics

65

Example 3.8 A closed and rigid cylinder has a volume of 0.34 m3 and contains helium

initially at 25◦ C and 6.0 × 105 Pa. More helium is added until the temperature reaches 75◦ C and the pressure reaches 20 × 105 Pa. The contents of the cylinder are then allowed to cool to the original temperature. Determine (a) the mass of helium added, (b) the final pressure, and (c) the heat lost during the cooling process.

Solution Assumptions and Specifications (1) The helium is the system and is an ideal gas. (2) “Closed and rigid” implies a constant volume process. (3) Kinetic and potential energy changes are negligible. (4) The helium addition and cooling processes occur under quasi-equilibrium conditions. (5) The specific heat may be taken from Table A.1 at 300 K. (a) We call the initial condition state 1 and with V1 = 0.34 m3 , P1 = 6.5 × 105 N/m2 and T1 = 25◦ C + 273◦ C = 298 K, we use Table A.1 to obtain for helium R = 2078 J/kg-K

and

c v = 3.146 kJ/kg-K

The original mass of helium in the tank can be found from the ideal gas law m1 =

P1 V1 (6 × 105 N/m2 )(0.34 m3 ) = = 0.329 kg RT1 (2078 J/kg-K)(298 K)

At state 2, V2 = 0.34 m3 and with P2 = 20 × 105 N/m2 and T2 = 75◦ C + 273◦ C = 348 K, the mass of the helium at state 2 will be m2 =

P2 V2 (20 × 105 N/m2 )(0.34 m3 ) = = 0.940 kg RT2 (2078 J/kg-K)(348 K)

This means that the mass of helium added between state 1 and state 2 is m = m2 − m1 = 0.940 kg − 0.329 kg = 0.611 kg ⇐ (b) We designate the final condition as state 3 where m3 = 0.940 kg, T3 = 298 K and V3 = 0.34 m3 . Thus, P3 =

m3 RT3 (0.940 kg)(2078 J/kg-K)(298 K) = = 1.712 × 106 Pa ⇐ V3 0.34 m3

(c) For this constant volume process with W = 0, Equations 3.12 and 3.31c give with T3 − T2 = 298 K − 348 K = − 50 K Q = m(u3 − u2 ) = mc v (T3 − T2 ) = (0.940 kg)(3.146 kJ/kg-K)(−50 K) = −147.9 kJ ⇐ The minus sign shows that heat is rejected to the surroundings.

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Introduction to Thermal and Fluid Engineering

3.10

Summary

The conservation of mass principle can be stated as: Mass can neither be created nor destroyed and its composition cannot be altered from one form to another unless it undergoes a chemical change. There are three forms of energy •

Kinetic energy KE =

•

m Vˆ 2

2

(N-m or joules)

Potential energy PE = mgz

•

(N-m or joules)

Internal energy, U, represents all energy changes in a system other than kinetic or potential energy. As specific internal energy, it is written as u=

U m

(N-m/kg or J/kg)

Total energy is the sum of kinetic, potential, and internal energy E = KE + PE + U Work is defined as the energy that is expended when a force acts through a displacement. The thermodynamic view of work is that it is a form of energy transfer between a system and its surroundings.  done by a system is positive Work −→ done on a system is negative and with regard to a system  W=

V2

Pd V V1

Other forms of work such as paddle wheel work, electrical work, and work required to elongate or compress a spring must often be considered. All energy transfer, other than work, across the boundaries of a system is accounted for by the heat transfer, Q. This heat transfer is induced by a temperature difference between the system and the surroundings and flows in the direction of decreasing temperature. Heat transfer is taken as positive when it is transferred to the system and negative when it is transferred from the system  transferred to a system is positive Heat −→ transferred from a system is negative From this, we note that the sign convection pertaining to heat transfer is exactly opposite to the sign convention adopted for the work.

Energy and the First Law of Thermodynamics

67

The first law of thermodynamics states that The change of total energy (kinetic, potential, and internal) is equal to the net heat transferred to the control mass minus the work done by the control mass. It may be considered as an alternative statement to the conservation of energy principle: Energy may neither be created nor destroyed but may be converted from one form to another. For a closed system where there is a change of kinetic or potential energy, the first law may be summarized by Equation 3.12 E2 − E1 = Q − W

(3.12)

The efficiency of a power cycle is =

Wcycle Qin − Qout Qout = =1− Qin Qin Qin

(3.18)

and for a refrigeration cycle, the coefficient of performance takes two forms. For the refrigerator Qin Qin = = (3.19) Wcycle Qout − Qin and for the heat pump =

Qout Qout = Wcycle Qout − Qin

In the ideal gas law PV = mRT

or

Pv = RT

the gas constant, R, depends on the universal gas constant R=

R¯ M

where R¯ = 8314 J/kgmol and M is the molecular weight of the gas. The enthalpy of a substance is H = U + PV

or

h = u + Pv

and the specific heats are defined by  du  cv = dT v = constant for constant volume and

 dh  cp = dT  P = constant

for constant pressure. For gases, the ratio of the specific heats is k=

cp cv

(3.20)

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Introduction to Thermal and Fluid Engineering

and the specific heats are related by c p = cv + R Processes of an ideal gas depend on P Vn = C

(a constant)

where the exponent, n, can vary. Three processes of an ideal gas are: •

The isometric or constant volume process where n = ∞

•

The isobaric or constant pressure process where n = 0

•

The isothermal or constant temperature process where n = 1

3.11

Problems

Forms of Energy and Work 3.1: A 16-kg mass has a potential energy of 4500 J with respect to a certain datum and g = 9.807 m/s2 . Determine the height relative to the datum. 3.2: A 200-N pile driver hammer is released 4 m above the top of a piling. For the pile driver hammer, determine, at the instant of impact, (a) the maximum change in potential energy and (b) the velocity if friction is negligible and g = 9.81 m/s.2 3.3: It takes 40 HP to drive an elevator system that has a mass of 3000 kg. Consider that the system as ideal, that is, with negligible losses. With g = 9.81 m/s,2 determine (a) the upward uniform velocity and (b) the kinetic energy. 3.4: Consider an elevator weighing 6000 N and a counterweight weighing 4800 N in motion with a velocity of Vˆ = 4.25 m/s. The system also contains a braking pulley and the kinetic energy of all cables and rotating parts is negligible. Determine the energy absorbed by the braking pulley if the elevator is stopped in 1.625 m and the gravitational acceleration is 9.81 m/s.2 3.5: A mass of 1500 kg is accelerated uniformly along a horizontal surface from 10.8 km/h to 10.8 km/h in 4 seconds. Determine (a) the change in kinetic energy, (b) the force required, and (c) the work done by this force over the time interval. 3.6: A 6-gm mass is dropped from rest from a height of 12 m above the earth’s surface. What will be its speed upon impact? 3.7: An elevator cab of weight W = 981 N moves from street level to the top of a tower, 412 m above the surface. Determine the change in potential energy. 3.8: Determine the energy expended by a climber weighing 600 N in climbing Mount Everest whose altitude is 8851 m. 3.9: The rolling resistance of a railroad car is 10 N per 6250 N of weight of car. Starting from rest, the car rolls down an incline 120 m long on a 5% grade. It then continues along a horizontal stretch. How far will the car roll on the horizontal stretch before coming to rest? 3.10: A mass of 4 kg moves at a velocity of 5 m/s at a height of 40 m above sea level. Determine (a) its weight, (b) its kinetic energy, and (c) its potential energy. 3.11: A disgruntled New York Mets fan drops a baseball with a mass of 143 g from a window in the Empire State Building at an elevation of 305 m. The ball reaches a terminal

Energy and the First Law of Thermodynamics

69

velocity of 40 m/s at the instant that it strikes the ground. Determine the increase in internal energy. 3.12: An object having a mass of 12 kg with an initial downward velocity of 35 m/s falls from a height of 125 m. The acceleration of gravity is 9.78 m/s2 and the air resistance is negligible. Determine the velocity at impact with the ground. 3.13: A 2-kg piece of metal is moving horizontally at 8 m/s. If g = 9.81 m/s2 , determine (a) the change in velocity necessary for an increase of kinetic energy of 12 N-m and (b) the change in elevation for an increase of potential energy of 16 N-m. 3.14: The acceleration of gravity varies as a function of the elevation above sea level in accordance with g = 9.807 m/s2 − 3.32 × 10−6 s−2 z where z is in meters. A satellite with a mass of 300 kg is boosted to an altitude of 450 km above the surface of the earth. Determine the work required. 3.15: A mass of 10 kg is 200 m above a given datum where g = 9.75 m/s2 . Determine the gravitational force in newtons and the potential energy in joules. 3.16: The work required to stretch a spring 8.59 cm from its free length is 362 J. The elemental equation that pertains to the spring is F = ks y where ks is the spring constant in N/m that relates the perturbing force, F in N to the stretch or compression of the spring, y in meters. Determine the value of the spring constant. 3.17: A weight of 480 N is attached on to the end of a rope 6 m below a frictionless pivot. The weight is pushed so that the rope makes an angle of 36.87◦ from the vertical. Determine (a) the gain in potential energy and (b) the mass if g = 9.81 m/s2 . 3.18: A quantity of gas at a constant pressure of 16 Pa is compressed from 6 m3 to 1.5 m3 . Determine the work done. 3.19: A sample of gas is expanded in a process at a constant pressure, P, from a volume of 1 m3 to a volume of 4 m3 . The amount of work done is 432 J. Determine the pressure. 3.20: Work is being done by a substance in a non-flow process in accordance with P=

100 V

(N/m2 )

where V is in m3 . Determine the work done on or by the substance as the pressure increases from 0.8 to 8 bar. The Energy Balance for Closed Systems 3.21: Consider a nonflow process from state point 1 to state point 2 where 100 J of work is done on the system and where 50 J of heat is transferred from the system. The initial internal energy is 50 J and potential and kinetic energy changes are negligible. Determine the final internal energy. 3.22: Consider a nonflow process from state point 1 to state point 2 where 100 J of work is done by the system and where 100 J of heat is transferred from the system. There is no change in internal energy and the increase in potential energy is 100 J. Determine the change in kinetic energy.

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Introduction to Thermal and Fluid Engineering

3.23: Consider a nonflow process from state point 1 to state point 2 where 100 J of work is done on the system and where 100 J of heat is transferred to the system. The initial internal energy is 80 J, the potential energy change is negligible and the kinetic energy increases by 40 J. Determine the final internal energy. 3.24: Consider a nonflow process from state point 1 to state point 2 where 100 J of heat is transferred to the environment, the internal energy increases by 20 J, the potential energy increases by 40 J, and the kinetic energy decreases by 40 J. Determine the magnitude and direction of the work done. 3.25: Consider a nonflow process from state point 1 to state point 2 where 20 J of work is done by the system and where 100 J of heat is transferred from the system. The final internal energy is 50 J, the potential energy increases by 40 J and the kinetic energy increases by 20 J. Determine the initial internal energy. 3.26: Consider a nonflow process from state point 1 to state point 2 where 50 J of work is done by the system. The change in internal energy is 100 J, the potential energy increases by 40 J and the change in kinetic energy is negligible. Determine the heat transferred. 3.27: Consider a nonflow process from state point 1 to state point 2 where 40 J of work is done by the system and where 40 J of heat is transferred to the system. The initial internal energy is 80 J and there is no increase in either potential or kinetic energy. Determine the final internal energy. 3.28: Consider a nonflow process from state point 1 to state point 2 where 40 J of work is done by the system and where there 100 J of heat is transferred from the system. The change in initial internal energy is 40 J and the kinetic energy increases by 60 J. Determine the change in potential energy. 3.29: Consider a nonflow process from state point 1 to state point 2 where 100 J of work is transferred to the environment, the internal energy increases by 50 J, the potential energy increase is negligible and no heat is transferred from or to the environment. Determine the change in kinetic energy. 3.30: Consider a nonflow process from state point 1 to state 2 where 200 J of heat is transferred to the system. Determine the work done if the internal energy decreases by 50 J, the potential energy increases by 50 J, and the kinetic energy increases by 50 J. 3.31: Consider a nonflow process from state point 1 to state point 2 where 40 kJ of work is done by the system and where 25 kJ of heat is transferred from the system. The system is returned to state 1, and the heat transferred to the system is 20 J. Determine the work done in the return process. 3.32: Consider a nonflow process from state point 1 to state point 2 where 50 kJ of work is done on the system and 100 kJ of heat is transferred to the system. The system is returned to state 1 and the heat transferred from the system is 20 J. Determine the work done in the return process. 3.33: Consider a nonflow process from state 1 to state 2 where 100 kJ of work is done by the system and no heat is transferred from or to the system. The system is returned to state 1 and the work done on the system is 80 kJ. Determine the heat transferred in the return process. 3.34: Consider a nonflow process from state 1 to state 2 where 12 kJ of work is done by the system and where 40 kJ of heat is transferred to the system. The process continues to state 3 with a heat transfer of 18 kJ to the system. During the return to state 1, 24 kJ of work is done by the system and the change of energy is E 31 = 12 kJ. Determine (a) Q31 , (b) W23 , (c) E 12 , and (d) E 23 .

Energy and the First Law of Thermodynamics

71

3.35: Consider a nonflow process from state 1 to state 2 where 50 kJ of work is done on the system and 80 kJ of heat is transferred to the system. The process continues to state 3 with a heat transfer of 25 kJ from the system. During the return to state 1, 10 kJ of work is done on the system and the heat transferred to the system is 35 kJ. Find (a) E 12 , (b) E 23 , (c) E 31 , and (d) W23 . The Ideal Gas 3.36: In a constant volume nonflow process, 2 kg of oxygen is heated from 30◦ C until the pressure is tripled. Treating oxygen as an ideal gas, determine (a) its final temperature and (b) the work done by the oxygen. 3.37: In a nonflow process 10 kg of air at 25◦ C and 1 bar are heated at constant pressure until the volume is 18 m3 . Treating air as an ideal gas and assuming constant specific heats, find (a) the change in internal energy, (b) the change in enthalpy, (c) the work done, and (d) the heat added. 3.38: A tank contains air at 50 bar and 25◦ C. The air may be assumed as an ideal gas and an amount measuring 20 m3 at 1 bar and 20◦ C is removed. The pressure in the tank is then found to be 20 bar when the temperature is 25◦ C. Determine the volume of the tank. 3.39: The temperature of an ideal gas remains constant while the pressure increases from 1 to 4 bar. The initial volume is 2.88 m3 . Determine (a) the final volume and (b) the work done. 3.40: An automobile tire contains a volume of air at a gage pressure of 2 bar and 24◦ C. The barometric pressure is 29.75 in of mercury. Under running conditions, the temperature of the air in the tire increases to 76◦ C. Determine the gage pressure. 3.41: An automobile tire has a volume of 0.052 m3 and contains air at a pressure of 2 bar and 25◦ C. Determine the mass of air in the tire. 3.42: For the automobile tire in Problem 3.41, determine the pressure in the tire if the automobile is left out so that the temperature falls to 0◦ C. 3.43: For the automobile tire in Problem 3.42, how many kilograms of air must be added to bring the pressure to 2.18 bar when the air temperature is 0◦ C? 3.44: Determine the density of oxygen at atmospheric pressure and 35◦ C. 3.45: At 37◦ C, 20 kg of propane occupies a volume of 3.125 m3 . Determine (a) the pressure and (b) the mass of propane that must be added to bring the pressure to 1.20 MPa. 3.46: Ethane contained in a 0.625 m3 tank at an absolute pressure of 225 kPa and 77◦ C begins to escape: 0.25 kg of the ethane leaks out of the tank and the temperature falls to 27◦ C. Determine the gage pressure of the ethane remaining in the tank if the barometer is at 99 kPa. 3.47: A balloon is filled with helium at 25◦ C and 30.12 in of mercury until the volume becomes 1000 m3 . Determine (a) the mass of helium required and (b) the volume if the balloon rises to an altitude where P = 28 in of mercury and the temperature is 0◦ C. 3.48: Suppose that hydrogen is substituted for helium in Problem 3.47. All other conditions remain the same. Determine (a) the mass of hydrogen required and (b) the volume if the balloon rises to an altitude where P = 28 in of mercury and the temperature is 0◦ C. 3.49: A closed rigid container with a volume of 4 m3 contains a gas having a molecular weight of 32 kg/kgmol at 1 MPa and 57◦ C. The container develops a leak and gas is

72

Introduction to Thermal and Fluid Engineering lost leaving a pressure of 0.4 MPa and a temperature of 27◦ C. Determine the volume of gas lost if it is reckoned at a pressure of 1 bar and a temperature of 25◦ C.

3.50: A rigid container weighing 200 N contains propane at 1.25 MPa and 27◦ C. After some propane has been used, the contents of the container are at 5.125 bar and 0◦ C and the container was then found to weigh 140 N. Determine (a) the mass of the container, (b) the internal volume of the propane container, and (c) the kilograms of propane used.

4 Properties of Pure, Simple Compressible Substances

Chapter Objectives •

To introduce and describe the state postulate.

To present several relationships that relate the pressure, volume, and temperature of a pure substance. • To develop thermodynamic property data including some hints for phase determination. • To investigate real gas behavior. •

•

To consider the principle of corresponding states.

•

To examine equations of state.

•

To provide data for the polytropic process of an ideal gas.

4.1

The State Postulate

We recall that a pure substance is one that is uniform and homogeneous in its chemical structure. It may exist in a single phase, or, if each phase is separated by a phase boundary, it may exist as a two- or three-phase mixture. However, a mixture of liquid water and an air-water vapor mixture is not a pure substance even though all its components remain in the gaseous phase (Figure 4.1). When a system undergoes a change of state, a large number of properties may change and, because intrinsic properties are those that are functions of molecular behavior, we expect that all intrinsic properties are functionally related. For example, suppose that the dependent intrinsic property, zo , is a function of p independent intrinsic properties, z1 , z2 , z3 , . . ., z p . Then zo = f (z1 , z2 , z3 , . . ., z p ) and once the values of z1 , z2 , z3 , . . ., z p are specified, the value of zo becomes fixed. However, we must be able to answer the question, “what is the value of p?” A simple compressible substance is defined as any pure substance where nonrelevant effects such as surface tension, velocity or motion, electrical and thermal conductivity, modulus of elasticity, and viscosity can be neglected. The words “simple compressible” imply that the only possible work interaction between the substance acting as the system and the surroundings that can change the state of the system is Pdv work. We note that, in 73

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Introduction to Thermal and Fluid Engineering

Air and Water Vapor

Water

FIGURE 4.1 A mixture of water and air containing water is not a pure substance.

general, for this case, the use of the specific volume and other specific quantities, permits the extent of the system to be removed from consideration. For a pure, simple compressible, substance, repeated observations have shown that two independent properties are necessary and sufficient to provide the equilibrium state of a system. While this observation cannot be derived from the basic postulates of the classical approach to thermodynamics, no violations have been observed. Thus, in the ideal gas model for a particular gas with a specified value of R, two properties specify a third P = f 1 (v, T),

v = f 2 ( P, T),

and

T = f 3 ( P, v)

The state postulate supplies the number of variables that are required to completely specify the state of a single-phase or homogeneous system. In general If there are n relevant work interactions between a system and its surroundings, the number of independent intensive and intrinsic properties required to specify the state of the system is n + 1. In particular, in the case of a simple compressible substance where Pdv is the only relevant work interaction, n = 1 and it is necessary to fix two independent properties to completely specify the equilibrium state.

4.2 P-v-T Relationships 4.2.1 Introduction The state postulate asserts that if P, v, and T are the properties of a simple compressible substance, then, for a single phase system, the pressure, P can be considered a function of the specific volume, v, and the temperature, T P = f (v, T)

(4.1)

The graph of such a function is called a surface and, in particular, the functional relationship of Equation 4.1 is a P-v-T surface. 4.2.2 The P -v-T Surface Figure 4.2 shows a P-v-T surface for a substance that expands upon freezing. Water is a typical example of such a substance. However, most substances contract upon freezing and the P-v-T surface for these substances is shown in Figure 4.3. We note that the coordinates of the P-v-T surface are pressure, specific volume, and temperature.

Properties of Pure, Simple Compressible Substances

75

Liquid

Pressure

Critical Point Solid

Sp ec

Liq uid Tri Va ple po L in r e So lid -

ific Vo lu

vap

Va po r

or

TC

p er Tem

me

re atu

FIGURE 4.2 The pressure-specific volume-temperature surface for a substance that expands upon freezing.

In Figures 4.2 and 4.3, we can see that there are three regions that contain a single phase. These are the solid, liquid, and vapor phases. Three two-phase regions are located between the single-phase regions and these are called the solid-liquid, the liquid-vapor, and the solid-vapor regions. 4.2.3 Single-Phase Regions

Solid-Liqu

id

In the single-phase regions of Figures 4.2 and 4.3, the state is fixed by any two of the properties, P, v, or T because all of these are independent.

Pressure

Solid

Liquid Critical Point

Liq Va uid po r Tri ple L in e So Sp ec

ific

Vo lu

me

r TC

lid

-va po

Va po

ConstantPressure Line

r

e tur era p Tem

FIGURE 4.3 The pressure-specific volume-temperature surface for a substance that contracts upon freezing.

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Introduction to Thermal and Fluid Engineering

4.2.4 Two-Phase Regions In the two-phase regions of Figures 4.2 and 4.3, two phases coexist at equilibrium. These are the solid-liquid, the liquid-vapor and the solid-vapor regions. In these two-phase regions, pressure and temperature are not independent and the state is fixed by specific volume and pressure or temperature. The change of phase from solid to liquid is referred to as melting and the change from liquid to solid is called freezing or solidification. The change from liquid to vapor is called vaporization or boiling and the change from vapor to liquid is called condensation. The change of phase from solid to vapor is known as sublimation. The quantity of heat required to change a solid to a liquid is called the latent heat of fusion or merely the heat of fusion. Similarly, the quantity of heat required to change a liquid to a vapor is called the latent heat of vaporization or merely the heat of vaporization. The quantity of heat required to change a solid to a vapor is called the heat of sublimation. The magnitudes of the latent heats depends upon the temperature at which the phase change occurs. For water at 0◦ C (273 K), the latent heat of fusion is f = 333.7 kJ/kg and at 100◦ C (373 K), the latent heat of vaporization is v = 2257.1 kJ/kg 4.2.5 Three-Phase Regions Three phases can coexist in equilibrium along a line shown in Figures 4.2 and 4.3, which is called the triple line and represents the triple state. For water, this state is found at the triple point where the pressure is 611.2 Pa and the temperature is 273.16 K (0.16◦ C). We may recall that the triple point of water was used to establish the Kelvin temperature scale. 4.2.6 The P -T Surface The projection of the P-v-T surface normal to the v-axis is called the P-T surface or phase diagram. The phase diagram for a substance that expands upon freezing is shown in Figure 4.4. The triple state is represented by a line on the P-v-T surface which is, in reality, a

Pressure

S

L

Solid

Critical Point

Liquid L V

S

Triple Point V Temperature

FIGURE 4.4 Phase diagram for water (not to scale).

Vapor

Properties of Pure, Simple Compressible Substances

77

Critical Point

Pressure

Saturated Liquid Line Saturated Vapor Line Compressed Liquid Region

LiquidVapor Region

Superheated Vapor Region

Specific Volume FIGURE 4.5 The P-v surface for water showing the saturated liquid and saturated vapor lines and the critical point.

line that proceeds into the plane of the paper and leaves at a point on the P-T diagram. This is the reason for designating the triple point as the point where all of the lines intersect. The lines between the phases are called saturation lines. We may note, in particular, the saturation line between the liquid and vapor planes that begins at the triple point and ends at a point called the critical point having a pressure designated as Pc and a temperature designated as Tc . The horizontal and vertical lines from this point are called the critical pressure isobars and the critical temperature isotherms, respectively. 4.2.7 The P -v Surface Water, a substance that expands upon freezing, can be used to provide a typical illustration of the P-v surface shown in Figure 4.5. Here, we see two saturation lines. The line that divides the liquid region, usually referred to as the compressed liquid or subcooled liquid region and the liquid-vapor region, is called the saturated liquid line. A compressed liquid is a liquid at a particular temperature whose pressure is higher than the saturation pressure at that temperature or liquid at a particular pressure whose temperature is lower than the saturation temperature at that pressure. Properties associated with the saturated liquid line are given the subscript f ( f for “fluid”) and typical examples are the specific volume, v f , the specific internal energy, u f and the specific enthalpy, h f . The line that divides the liquid-vapor region and the vapor region, called the superheated vapor region, is called the saturated vapor line. Properties associated with the saturated vapor are given the subscript g (g for “gas”) and typical examples are the specific volume, vg , the specific internal energy, ug and the specific enthalpy, h g . The difference in properties between the saturated liquid and saturated vapor lines are given the subscript f g (for “the difference between gas and fluid”) and typical examples are ufg and h fg . In Figure 4.5, we can observe that the saturated liquid and saturated vapor lines intersect at the critical point and form a vapor dome. The area under the vapor dome represents the so-called wet region because the liquid-vapor region contains both liquid and vapor to varying degrees. Thermodynamic properties under the vapor dome (with the saturated liquid and saturated vapor lines as extremes) depend on a quantity known as the quality that is the mass fraction of the vapor present. x=

mvap mg = mtot m f + mg

(4.2)

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Introduction to Thermal and Fluid Engineering

where the subscripts represent the amounts of liquid and vapor present. Sometimes, y, the amount of liquid or moisture present, is employed y=

mliq mf = mtot m f + mg

(4.3)

Of course, there is a relationship between x and y x+y=1 4.2.8 Some Hints For Phase Determination The location of the phase of a substance on the P-v surface can be facilitated by noting that the saturation pressure and the saturation temperature are coupled. •

To be in the superheated vapor region, the temperature must be greater than the saturation temperature.

•

To be in the liquid-vapor region, the temperature must be equal to the saturation temperature and the pressure must be equal to the saturation pressure.

•

To be in the subcooled or compressed liquid region, the temperature must be less than the saturation temperature.

4.3

Thermodynamic Property Data

4.3.1 Table Arrangement Thermodynamic property data are available in listed equations, graphs, computer codes, or tables such as those published by Keenan et al. (1969) and ASME (1967). In the steam tables, extractions of which are included here in Tables A.3 to A.5 in Appendix A, the data are divided into convenient increments and linear interpolation may be employed to provide results to an acceptable accuracy. Tables for Refrigerant R-134a (a relatively new non-CFC refrigerant) extracted from ASHRAE (1994) are given in the same format as the steam tables in Tables A.6 to A.8 in Appendix A. The tables are categorized into: •

Saturation tables (Tables A.3 and A.4 for water-steam and Tables A.6 and A.7 for refrigerant R-134a),

•

Superheated vapor tables (Table A.5 for steam and Table A.8 for refrigerant R-134a), and

•

Compressed liquid tables.

Details for the saturation and superheated vapor tables will be considered in Sections 4.3.2 and 4.3.3, which now follow, and an alternative method for the evaluation of compressed liquid properties is provided in Section 4.3.4. 4.3.2 The Saturation Tables The saturation tables provide values of specific volume, v, specific internal energy, u, specific enthalpy, h, and specific entropy, s, a property that will be introduced to our study in

Properties of Pure, Simple Compressible Substances

79

Pressure and Temperature

P, T

vf, uf, hf, sf vg, ug, hg, sg

ufg, hfg, sfg

vx, ux, hx, sx Point at Quality, x v

Specific Volume FIGURE 4.6 The basis for the saturation tables.

Chapter 7. The properties are listed for both the saturated liquid and saturated vapor states that form the boundaries of the vapor dome and are to be used for the liquid-vapor region where either the pressure, P, or the temperature, T, must be provided. Table A.3 is the temperature table where the saturation pressure is also given to the immediate right of the saturation temperature. Table A.4 is the pressure table where the saturation temperature is listed to the right of the saturation pressure. We recall that the subscripts f and g, respectively, designate a property of a saturated liquid and a saturated vapor. The values that designate the difference between saturated vapor and saturated liquid are designated by the subscript fg. At points under the vapor dome that are neither saturated liquid nor saturated vapor, the property value is obtained by adding the contributions of the two phases. For example, with the quality and moisture fraction defined by Equations 4.2 and 4.3, we find that the volume in the liquid-vapor region having a quality, x, will be V = Vliq + Vvap and with the specific volume at point x designated as vx vx =

Vliq Vvap + m m

and then mvx = m f v f + mg vg Because m f = m − mg mvx = (m − mg )v f + mg vg and then a division by the total mass provides  mg  mg vx = 1 − vf + vg = (1 − x)v f + xvg m m

(4.4a)

80

Introduction to Thermal and Fluid Engineering If the difference between vg and vf is vfg = vg − v f

then Equation 4.4a can be rearranged to vx = v f + xvfg

(4.4b)

In those cases where values of vfg are not listed, Equation 4.4a may be employed. In the cases of specific internal energy, specific enthalpy and specific entropy, which is introduced in Chapter 7, ux = (1 − x)u f + xug

(4.5a)

ux = u f + xufg

(4.5b)

h x = (1 − x)h f + xh g

(4.6a)

h x = h f + xh fg

(4.6b)

sx = (1 − x)s f + xsg

(4.7a)

sx = s f + xsfg

(4.7b)

or

or

and

or

If the property of a substance in the liquid-vapor region is given, the quality can be found via x=

vx − v f ux − u f hx − h f sx − s f = = = vg − v f ufg h fg sfg

(4.8)

Here, vx , ux , h x , or sx is the given property.

Example 4.1 At 160◦ C, 5 kg of water are enclosed in a rigid container having a volume of 0.5 m3 (Figure 4.7). Determine (a) the pressure, (b) the enthalpy, and (c) the mass and volume of the vapor.

V = 0.5 m3

5 kg of water at 160°C

FIGURE 4.7 The control volume for Example 4.1.

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81

Solution Assumptions and Specifications (1) The 5 kg of water is the system. (2) There is vapor present (this will require checking). (3) The given conditions of V and T represent an equilibrium state. The specific volume of the mixture will be vmix =

V 0.50 m3 = = 0.10 m3/kg m 5 kg

Because the temperature is given, we can find from Table A.3 that v f = 1.102 × 10−3 m3/kg

vg = 0.3068 m3/kg

and

We note that the mixture specific volume is between v f and vg . Thus, the mixture exists in the liquid-vapor phase and assumption 2 has been verified. (a) The pressure is the saturation pressure and we can read from Table A.3 at 160◦ C. Psat = 0.618 MPa

(618 kPa) ⇐

(b) The enthalpy value depends on the quality that we can find from the specific volume data. From Equation 4.8 x=

vx − v f vg − v f

=

0.1000 m3/kg − 1.102 × 10−3 m3/kg 0.3068 m3/kg − 1.102 × 10−3 m3/kg

=

0.0989 m3/kg = 0.324 0.3057 m3/kg

Then, with h f and h fg from Table A.3 h f = 675.2 kJ/kg

and

h fg = 2081.7 kJ/kg

the value of h x is found from Equation 4.6b h x = h f + xh fg = 675.2 kJ/kg + (0.324)(2081.7 kJ/kg) = 675.2 kJ/kg + 674.5 kJ/kg = 1349.7 kJ/kg ⇐ (c) The mass of vapor, mg , depends on the quality mg = xmmix = (0.324)(5 kg) = 1.620 kg ⇐ and the volume of the vapor will be Vg = mg vg = (1.620 kg)(0.3068 m3/kg) = 0.497 m3 ⇐ We note that the vapor occupies almost the entire container.

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x = 0.40

P = 300 kPa

Liquid Water

FIGURE 4.8 The control volume for Example 4.2.

Example 4.2 A mass of saturated liquid water is enclosed in a container where the pressure is held at 300 kPa (Figure 4.8). Heat is added to the water until the quality is 40%. Determine (a) the initial temperature, (b) the final temperature, (c) the final specific volume, and (d) the final internal energy.

Solution Assumptions and Specifications (1) The water and any vapor produced by the heat addition constitutes the system. (2) Both the initial state and the state after the heat addition represent equilibrium states. (a) At 300 kPa. Table A.4 gives a saturation temperature of 133.5◦ C ⇐ (b) A vaporization of some of the water to form a mixture at a certain quality merely shows that the mixture has moved at constant pressure and temperature from the saturated liquid line to a point somewhere under the vapor dome. Thus, the temperature remains at the saturation value 133.5◦ C ⇐ (c) With v f and vg taken from Table A.4 at 300 kPa v f = 1.073 × 10−3 m3/kg

and

vg = 0.6056 m3/kg

the final specific volume at a quality of 40% can be found from Equation 4.4a vx = (1 − x)v f + xvg = (1 − 0.40)(1.073 × 10−3 m3/kg) + (0.40)(0.6056 m3/kg) = 6.438 × 10−4 m3/kg + 0.2422 m3/kg = 0.2429 m3/kg ⇐ (d) With u f and ufg from Table A.4 u f = 560.83 kJ/kg

and

ufg = 1982.1 kJ/kg

the value of ux is found from Equation 4.5b ux = u f + xufg = 560.83 kJ/kg + (0.40)(1982.1 kJ/kg) = 560.83 kJ/kg + 792.84 kJ/kg = 1353.7 kJ/kg ⇐

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83

P

Pressure

x2 = x1 P2 v=C P1

120°C

x1

v

Specific Volume FIGURE 4.9 Pressure-volume diagram for Example 4.3.

The following example shows that, under certain conditions, the addition of heat can cause a liquid-vapor mixture to become saturated liquid.

Example 4.3 A closed rigid vessel contains water in the form of a liquid-vapor mixture at

120◦ C. The specific volume of the mixture is 0.002213 m3/kg. Heat is added to the mixture until all of the vapor condenses and the mixture turns into a saturated liquid. Determine (a) the initial pressure of the mixture, (b) the initial quality of the mixture, (c) the final pressure, (d) the final temperature, and (e) the heat added during the process.

Solution Assumptions and Specifications (1) The liquid vapor mixture constitutes the system. (2) Both the initial state and the state after the heat addition represent equilibrium states. (a) Because the initial state is a mixture of liquid and vapor, the initial pressure must be the saturation pressure corresponding to 120◦ C. Table A.3 gives the pressure as P1 = 198.55 kPa ⇐ (b) With v f and vg taken from Table A.3 at 120◦ C v f = 1.060 × 10−3 m3/kg

and

vg = 0.8921 m3/kg

we calculate the quality of the mixture at point 1 using Equation 4.8 x=

vx − v f vg − v f

=

2.213 × 10−3 m3/kg − 1.060 × 10−3 m3/kg 0.8921 m3/kg − 1.060 × 10−3 m3/kg

=

1.153 × 10−3 m3/kg = 0.00129 ⇐ 0.8910 m3/kg

This shows that at point 1, just a trace of vapor is present.

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(c) Because the container is closed and rigid, the final specific volume must equal the initial specific volume and we find that at point 2, v2 = 2.213 × 10−3 m3/kg The final pressure is found from Table A.4 with interpolation P2 = 21.012 MPa ⇐ (d) Noting that the final state lies on the saturated liquid line, the final temperature is read from Table A.4 with interpolation as T2 = 369.8◦ C ⇐ (e) Because no work is done in a constant volume process, the first law of thermodynamics gives Q = u2 − u1 m The initial internal energy is found from Table A.3 at 120◦ C. With u f = 503.28 kJ/kg

and

ufg = 2025.4 kJ/kg

the value of u1 is found from Equation 4.5b u1 = u f + xufg = 503.28 kJ/kg + (0.00129)(2025.4 kJ/kg) = 503.28 kJ/kg + 2.61 kJ/kg = 505.89 kJ/kg ⇐ and with u2 read from Table A.4 at 369.8◦ C u2 = u f = 1839.7 kJ/kg Hence ˙ Q = u2 − u1 m = 1839.7 kJ/kg − 505.89 kJ/kg = 1333.8 kJ/kg ⇐ The process is shown on a P-v diagram in Figure 4.10. It is interesting to note that if the specific volume at state 1 is less than vc (v1 < vc ), the constant volume heat addition results in a saturated liquid (state 2). However, if the initial specific volume at state 1 is greated than vc (v1 > vc ), the constant volume heat addition results in a saturated vapor (state 2 ).

4.3.3 The Superheated Vapor Tables In the single-phase superheat region, two properties are needed to specify the equilibrium state. The variables chosen are the temperature, T, and the pressure, P. Table A.5 gives values of v, u, h, and s at several locations that form a grid of values based on temperature

Properties of Pure, Simple Compressible Substances

85

Pressure or Temperature

P, T

2 2' 1'

1 v1 < vC

vC

v1 > vC v

Specific Volume FIGURE 4.10 Constant volume processes in a liquid-vapor mixture.

and pressure. For points that do not fall within the grid, linear interpolation provides reasonable and accurate values.

Example 4.4 Determine (a) the internal energy of steam at 1.00 MPa and 340◦ C and (b) the enthalpy of steam at 3.00 MPa and 264◦ C.

Solution Assumptions and Specifications (1) The steam constitutes a closed system. (2) The states represent equilibrium states. (a) We note in Table A.5, that a point for 1.00 MPa and 340◦ C is not provided. Thus, we must interpolate using values at 1.00 MPa and 300◦ C and 1.00 MPa and 350◦ C. From Table A.5 at 1.00 MPa and 350◦ C, read u = 2874.7 kJ/kg T

Critical Point

P1 > P2 > P3

Temperature

P1 P2

Superheated Vapor Region

P3 Saturated Vapor Line

Specific Volume FIGURE 4.11 The superheated vapor region.

v

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Introduction to Thermal and Fluid Engineering

and at 1.00 MPa and 300◦ C, read u = 2793.1 kJ/kg The value of the internal energy at 1.00 MPa and 340◦ C can now be found by linear interpolation. We establish an interpolation fraction, F , as F =

340◦ C − 300◦ C = 0.80 350◦ C − 300◦ C

and the value of u will be u = 2793.1 kJ/kg + F (2874.7 kJ/kg − 2793.1 kJ/kg) = 2793.1 kJ/kg + (0.80)(81.6 kJ/kg) = 2793.1 kJ/kg + 65.3 kJ/kg = 2858.4 kJ/kg ⇐ (b) This time, we note that in Table A.5, that a point for 3.00 MPa and 264◦ C is not provided. Thus, we must interpolate using values at 3.00 MPa and 250◦ C and 3.00 MPa and 300◦ C. From Table A.5 at 3.00 MPa and 300◦ C, read h = 2994.1 kJ/kg and at 3.00 MPa and 250◦ C, read h = 2854.9 kJ/kg Here, the interpolation fraction is F =

264◦ C − 250◦ C = 0.28 300◦ C − 250◦ C

and the value of the enthalpy at 3.00 MPa and 264◦ C will be h = 2854.9 kJ/kg + F (2994.1 kJ/kg − 2854.9 kJ/kg) = 2854.9 kJ/kg + (0.28)(139.2 kJ/kg) = 2854.9 kJ/kg + 39.0 kJ/kg = 2893.9 kJ/kg ⇐ 4.3.4 The Compressed or Subcooled Liquid Region Although a compressed liquid table is usually provided in compendia of thermodynamic properties, we provide a close approximation. This approximation is based upon the fact that that there is little variation in the values of v and u at pressures above saturation values. Thus, v( P, T) ≈ v(T)

(4.9a)

u( P, T) ≈ u(T)

(4.9b)

and

The approximate value of the enthalpy, h, in the compressed (subcooled) liquid region can be obtained for a given temperature, T, having an associated saturation pressure, Psat ,

Properties of Pure, Simple Compressible Substances

87

T Critical Point

Temperature

Compressed Liquid Region

P1

P1 > P2 > P3

P2 P3 P1 P2 P3 v Specific Volume

FIGURE 4.12 The compressed liquid region.

by noting that h( P, T) = u( P, T) + Pv( P, T) Then, with the approximations expressed in Equation 4.9, values of the enthalpy in the compressed liquid region may be evaluated according to the deviation from compressed liquid dh = du + d( Pv) h − h f = u − u f + ( Pv) − ( Pv) f = 0 + v f ( P − Pf ) or h( P, T) ≈ h f (T) + v f ( P − Psat )

(4.10)

where h( P, T) is the value of the enthalpy sought at the compressed liquid pressure, P.

Liquid Water at 100°C and 10 MPa

FIGURE 4.13 The control volume for Example 4.5.

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Introduction to Thermal and Fluid Engineering

Example 4.5 Determine the enthalpy of liquid water at a temperature of 100◦ C and a pressure of 10 MPa.

Solution Assumptions and Specifications (1) The liquid constitutes a closed system. (2) The liquid is in equilibrium state. We can use Table A.3 at 100◦ C to find Psat = 0.1013 MPa,

h f = 418.9 kJ/kg

and v f = 1.043 × 10−3 m3/kJ

Then with P − Psat = 10 MPa − 0.1013 MPa = 9.8987 MPa so that v f ( P − Psat ) = (1.043 × 10−3 m3/kg)(9.8987 MPa) = 10.32 kJ/kg and Equation 4.10 provides h ≈ h f + v f ( P − Psat ) ≈ 418.9 kJ/kg + 10.32kJ/kg ≈ 429.2kJ/kg ⇐

Example 4.6 Refrigerant R-134a which is at 8◦ C and a quality of x = 0.60 (Figure 4.14) is

enclosed in a closed and rigid container having a volume of 0.115 m3 . Heat is added until the pressure is 800 kPa. Determine the quantity of heat added.

Solution Assumptions and Specifications (1) The refrigerant and any vapor produced by the heat addition constitutes the system. (2) Both the initial state and the state after the heat addition represent equilibrium states. (3) Kinetic and potential energy changes are negligible. (4) A closed and rigid container implies a constant volume heat addition.

V = 0.115 m3 Refrigerant 134a at 8°C and x = 0.60

Q FIGURE 4.14 The control volume for Example 4.6.

Properties of Pure, Simple Compressible Substances

89

Here, because no work is done on or by the system (constant volume process) Q = U − W = U = m(u2 − u1 ) At the initial state designated by point 1, Table A.6 at 8◦ C gives v f = 7.884 × 10−4 m3/kg

and vg = 0.0525 m3/kg

and u f = 60.43 kJ/kg

and ug = 231.46 kJ/kg

At x = 0.60 using Equation 4.4a v1 = (1 − x)v f + xvg = (1 − 0.60)(7.884 × 10−4 m3/kg) + (0.60)(0.0525 m3/kg) = 3.154 × 10−4 m3/kg + 0.0315 m3/kg = 0.0318 m3/kg and with u1 as the internal energy at point 1 u1 = (1 − x)u f + xug = (0.40)(60.43 kJ/kg) + (0.60)(231.46 kJ/kg) = 24.17 kJ/kg + 138.88 kJ/kg = 163.05 kJ/kg With v = v1 , the mass of refrigerant R-134a will be m=

V 0.115 m3 = = 3.616 kg v 0.0318 m3/kg

We next observe in Table A.7 that, at 800 kPa or 0.8 MPa, the saturated vapor specific volume is vg = 0.0255 m3/kg so that the final state at point 2 is seen to be in the superheated vapor region. Thus, Table A.8 can be employed to provide at v2 = v1 = 0.0318 m3 /kg with interpolation T2 = 73.7◦ C The interpolation process also yields the internal energy at point 2 u2 = 283.74 kJ/kg The amount of heat added is Q = U − W = U or Q = m(u2 − u1 ) = (3.616 kg)(283.74 kJ/kg − 163.05 kJ/kg) = (3.616 kg)(120.69 kJ/kg) = 436.4 kJ ⇐

90

4.4

Introduction to Thermal and Fluid Engineering

The T-s and h-s Diagrams

To this point, we have relied on the P-v diagram to show thermodynamic properties for water and refigerant R-134a. Indeed, the P-v diagram (the P-v surface considered in Section 4.2.7 and Figure 4.5) has been employed to show the compressed liquid, the liquid-vapor, and the superheated vapor regions. The thermodynamic property of entropy is included in thermodynamic property data and while this property in introduced in Chapter 7, it is well, at this juncture, to describe two additional and extremely useful property diagrams containing specific entropy. Figure 4.15a displays the temperature-specific entropy (T-s) diagram where the critical point is located at the apex of the vapor dome and where the saturated liquid and saturated vapor lines flow downward to the left and right of the critical point. In the T-s diagram, constant pressure lines are horizontal inside of the vapor dome and constant specific enthalpy T

Lines of Constant Pressure

Critical Point

Sat ura ted

Liq

uid

Lines of Constant Enthalpy

s

(a)

Lines of Constant Pressure T

Sat ura ted

Vap or

Lines of Constant Temperature

Critical Point

s

(b) FIGURE 4.15 (a) The temperature-specific entropy (T-s) diagram and (b) specific enthalpy-specific entropy (h-s) diagram.

Properties of Pure, Simple Compressible Substances

91

lines become nearly horizontal in the superheated vapor region as the pressure is reduced. This is indicated in the shaded area in Figure 14.15a and for pressures and temperatures outside of the shaded area, both temperature and pressure are needed to evaluate the specific enthalpy. The specific enthalpy-specific entropy (h-s) diagram is shown in Figure 4.15b. Known as the Mollier diagram or Mollier chart, the diagram does not provide any liquid data. The critical point is located in the lower left-hand corner and the saturated vapor line may be noted. In the superheated vapor region, constant temperature lines become nearly horizontal as the pressure is reduced and this is indicated by the shaded area.

4.5

Real Gas Behavior

4.5.1 Introduction On a macroscopic scale, a gas appears as if it were a homogeneous system described by its pressure, volume, and temperature. However, an examination of a gas on a microscopic basis requires the consideration of a huge number of molecules in random motion and with considerable intermolecular space between them. For example, the pressure that a gas exerts on a surface is related to the number of molecules striking the surface. As the pressure increases or the temperature decreases, the mean free path between adjacent molecules decreases. The density will increase (or the specific volume will decrease) and the ideal gas relationship given in Section 3.7 by Equation 3.22 and repeated here as Pv = RT

(4.11)

will no longer be valid at very high pressures. We illustrate this in the next example.

Example 4.7 Determine the percent error between the specific volume to be obtained from the steam table and the value obtained from the ideal gas law for steam at (a) 15 MPa and 400◦ C and (b) 15 MPa and 600◦ C.

Solution Assumptions and Specifications (1) The steam is a closed system. (2) The steam is in the equilibrium state. (a) At 15 MPa and 400◦ C (673 K), Table A.5 gives v = 0.01566 m3/kg

Steam at 15 MPa & 400°C

Steam at 15 MPa & 600°C

(a)

(b)

FIGURE 4.16 Two alternative control masses for Example 4.7.

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For water vapor with a molecular weight of M = 18.02 kg/kgmol, we obtain the value of R from ¯ R 8.314 kJ/kgmol-K R= = = 0.4614 kJ/kg-K M 18.02 kg/kgmol so that we can obtain v from Equation 4.11 with T = 673 K v=

RT (0.4614 kN-m/kg-K)(673 K) = = 0.02070 m3/kg P 15 × 103 kN/m2

To establish the percent error, we take the steam table value as the base value because it is based on experimental evidence. Hence,   0.02070 m3/kg − 0.01566 m3/kg % error = 100 0.01566 m3/kg   0.00504 m3/kg = 100 0.01566 m3/kg = 32.19 % ⇐ (b) At 15 MPa and 600◦ C (873 K), Table A.3 gives v = 0.02488 m3/kg Then from Equation 4.11, we obtain v=

RT (0.4614 kPa-m3/kg-K)(873 K) = = 0.02685 M3/kg P 15 × 103 kPa

This time, the percent error is 

0.02685 m3/kg − 0.02448 m3/kg % error = 100 0.02448 m3/kg   0.00237 m3/kg = 100 0.02448 m3/kg



= 9.70 % ⇐ It appears that the error is greater as the saturated vapor line is approached. Moreover, we can see very clearly that there is a danger in considering water vapor as an ideal gas.

4.5.2 The Compressibility Factor Gases do not follow the ideal gas equation unless the pressure is relatively low and the temperature is relatively high. We therefore need a method for correcting the ideal gas equation to allow for real gas behavior and still maintain reasonable accuracy. The ideal gas law of Equation 4.11 may be used to define a compressibility factor, Z Z=

Pv RT

(4.12)

and this indicates that a gas can be taken as an ideal gas only if Z=

Pv =1 RT

(4.13)

Properties of Pure, Simple Compressible Substances

93

The use of the compressibility factor, Z, permits the prediction of P-v-T data when a gas is no longer at low pressure and moderate temperatures. Values of Z for a particular gas can be developed on the basis of Equation 4.13 for a wide range of equilibrium states and may be plotted for a particular gas as functions of pressure and temperature. When the pressure and temperature are normalized with respect to their values at the critical state, all gases can be seen to fit approximately on a single compressibility chart. Critical state properties are provided in Table A.9 and the normalizations involve three reduced properties. The reduced pressure is Pr ≡

P Pc

(4.14)

where both P and Pc must be in consistent units to make Pr dimensionless. The reduced temperature is Tr ≡

T Tc

(4.15)

with both temperatures in absolute units. Because the use of the reduced specific volume presents computational difficulties, it has been replaced by the pseudo-reduced specific volume v vr ≡ (4.16) RTc /Pc Because the ideal gas specific volume is vid =

RT P

we may consider Z as the ratio of the actual gas specific volume to the ideal gas specific volume v Z= vid 4.5.3 The Principle of Corresponding States The principle of corresponding states asserts that The compressibility factor for all gases is approximately the same when gases have identical reduced pressure and temperature. Thus, the compressibility factor for any gas is a function of the reduced pressure and temperature Z = f ( Pr , Tr )

(4.17)

and the generalized compressibility charts with values of Pr plotted as the abscissa and Z plotted as the ordinate are displayed in Figures A.1 through A.3 in Appendix A.

Example 4.8 In Example 4.7, the specific volume of steam, obtained from the ideal gas equation, was seen to differ markedly from the steam table value. Repeat Example 4.7 using the principle of corresponding states.

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Solution Assumptions and Specifications (1) The steam is a closed system. (2) The steam is in the equilibrium state. Example 4.7 gives R = 0.4641 kJ/kg-K and, from Table A.9, we find for water Pc = 22.09 MPa

and

Tc = 647.3 K

Then, for P = 15 MPa Pr =

P 15 MPa = = 0.679 Pc 22.09 MPa

(a) For T = 673 K Tr =

T 673 K = = 1.040 Tc 647.3 K

Figure A.3 gives Z = 0.755 so that v=

ZRT (0.755)(0.4641 kN-m3/kg-K)(673 K) = = 0.01572 m3/kg P 15 × 103 kPa

With the steam table value of 0.01566 m3 /kg, we see that the error is now almost negligible. (b) For T = 873 K Tr =

T 873 K = = 1.349 Tc 647.3 K

Figure A.3 then gives Z = 0.91 so that v=

ZRT (0.91)(0.4641 Pa-m3/kg-K)(873 K) = = 0.02458 m3/kg P 15 × 103 kPa

With the steam table value of 0.02588 m3 /kg, we find that the error here is also negligible.

4.6

Equations of State

We are aware that the ideal gas law of Equation 4.11 may not always provide an accurate representation among pressure, specific volume, and temperature. More accurate relationships among these variables can be obtained by developing analytical formulations that are called equations of state. The generalized equations of state are usually equations that contain two constants and are not very complicated. These include

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95

The van der Waals equation • The Berthold equation •

•

The Redlich-Kwong equation

These equations have a limited range of application because of their simplicity and have been developed from generalized behavior at the critical state. The empirical equations of state include •

The Beattie-Bridgeman equation, which has five constants

•

The Benedict-Webb-Rubin equation, which has eight constants

These equations are more complex than the generalized equations of state. The constants have been determined from experimental data. A theoretical equation of state is the virial equation of state in which the product of the pressure and the molal specific volume is represented by an infinite series. The constants in the infinite series, called virial constants, are based on principles derived from the kinetic theory of gases. We will provide some details for the van der Waals and the Redlich-Kwong equations of state. 4.6.1 The Van der Waals Equation The van der Waals equation of state, proposed in 1873, is P=

¯ RT a − 2 v¯ − b v¯

(4.18)

¯ 2 Tc2 27 R 64 Pc

(4.19a)

where a=

attempts to correct for intermolecular forces and b=

¯ c RT 8Pc

(4.19b)

which is intended to account for the volume of the gas molecules. The term, v¯ , is the molal specific volume for the gas. The van der Waals equation is only accurate over a limited range with an accuracy that decreases at high densities. The van der Waal constants, a and b for seventeen gases are given in Table A.10. 4.6.2 The Redlich-Kwong Equation of State The Redlich-Kwong equation of state P=

¯ RT a − 1/2 v¯ − b T v¯ ( v¯ + b)

(4.20)

proposed in 1949, provides more accuracy than the van der Waals equation of state. In Equation 4.20 a=

¯ 2 Tc2.50 0.4275 R Pc

(4.21a)

96

Introduction to Thermal and Fluid Engineering V = 0.0375 m3 5 kg of CH4 at –60°C

FIGURE 4.17 The control volume for Example 4.9.

and b=

¯ c 0.0867 RT Pc

(4.21b)

The Redlich-Kwong constants, a and b for eight gases are given in Table A.11.

Example 4.9 As indicated in Figure 4.17, consider 5 kg of methane at −60◦ C in a closed and rigid container having a volume of 0.0375 m3 . Determine the pressure in bars exerted by the methane using (a) the ideal gas equation of state, (b) the principle of corresponding states, (c) the van der Waals equation of state, and (d) the Redlich-Kwong equation of state.

Solution Assumptions and Specifications (1) The methane in the chamber is a closed system. (2) Equilibrium exists. ¯ = 8314 N-m/kgmol-K and the molecular weight of methane, M = 16.04 kg/kgmol With R taken from Table A.1, we have     V 0.0375 m3 v¯ = Mv = M = (16.04 kg/kgmol) = 0.1203 m3/kgmol m 5 kg (a) By the ideal gas equation of state at T = −60◦ C or 213 K P=

¯ RT (8314 J/kgmol-K)(213 K) = = 147.21 bars ⇐ v¯ 0.1203 m3/kgmol

(b) By the method of corresponding states, Table A.9 gives for methane Tc = 191 K and

Pc = 46.4 bar

Then, Tr =

T 213 K = = 1.115 Tc 191 K

and because Pr is to be found, we need, vr as the second parameter in Figure A.2. Thus, vr =

v Pc v¯ Pc (0.1203 m3/kgmol)(46.4 × 105 N/m2 ) = = = 0.352 ¯ c RTc RT (8314 N-m/kgmol-K)(191 K)

Then we read Figure A.2 to find at Tr = 1.115 and vr = 0.352 either Z = 0.508 or

Pr = 1.59

Properties of Pure, Simple Compressible Substances

97

From Z we have P=

¯ Z RT (0.508)(8314 N-m/kgmol-K)(213 K) = = 74.78 bar v¯ 0.1203 m3/kgmol

and from Pr we have P = Pr Pc = (1.59)(46.4 bars) = 73.77 bar ⇐ The discrepancy here is due to the lack of precision in reading Figure A.2. (c) The constants in the van der Waals equation are obtained for methane from Table A.10 a = 2.293 bar-(m/kgmol) 2

and b = 0.0428 m3/kgmol

Then a 2.293 bar-(m/kgmol) 2 = = 158.44 bar v¯ 2 (0.1203 m3/kgmol) 2 and ¯ RT (8314 N-m/kgmol-K)(213 K) = = 228.50 bar v¯ − b 0.1203 m3/kgmol − 0.0428 m3/kgmol Thus, by Equation 4.18 we have P=

¯ RT a¯ − 2 = 228.50 bar − 158.44 bar = 70.06 bar ⇐ v¯ − b v¯

(d) The constants in the Redlich-Kwong equation are obtained for methane from Table A.11 a = 32.19 bar-(m/kgmol) 2 -K0.50

and

b = 0.02969 m3/kgmol

Then with v¯ ( v¯ + b) = 0.1203 m3/kgmol)(0.1203 m3/kgmol + 0.0297 m3/kgmol) or v¯ ( v¯ + b) = 0.0180 (m3/kgmol) 2 we have a 32.19 bar-(m3/kgmol) 2 -K0.50 = = 122.53 bar T 1/2 v¯ ( v¯ + b) (213 K) 0.50 (0.0180 m3/kgmol2 ) and ¯ RT (8314 N-m/kgmol-K)(213 K) = = 195.44 bar v¯ − b 0.1203 m3/kgmol − 0.02969 m3/kgmol Thus, by Equation 4.20 we have P=

¯ RT a − 1/2 = 195.44 bar − 122.53 bar = 72.91 bar ⇐ v¯ − b T v¯ ( v¯ + b)

98

Introduction to Thermal and Fluid Engineering We may compare the foregoing values: Method

Pressure, bars

Ideal gas law Corresponding states van der Waals equation Redlich-Kwong equation

147.21 74.78 70.06 73.91

We observe that the method of corresponding states and the Redlich-Kwong equation provide almost the same value of the pressure. The ideal gas equation is, clearly, inaccurate at this relatively high pressure. The Redlich-Kwong equation of state of 1949 is more accurate than the van der Waals equation of state developed in 1873. Hence, we are led to the conclusion that the pressure is approximately 73.50 bar.

4.7

The Polytropic Process for an Ideal Gas

In Section 3.9, we suggested that processes involving ideal gases can be characterized by Equation 3.36 P Vn = C

(3.36)

where the exponent, n, may have values that are dictated by the particular process. Pertinent relationships were then developed for the constant volume process (n = ∞), the constant pressure process (n = 0), and the the constant temperature process (n = 1). For the polytropic process, n may take on any value from −∞ to ∞ (∞ < n < ∞). 4.7.1

P -V-T Relationships

In the polytropic process beginning at state 1 and ending at state 2 P1 V1n = P2 V2n = C n may range between −∞ < n < ∞. The P-V-T relationship is P2 = P1



V1 V2

n (4.22)

and because P1 V1 /T1 = P2 V2 /T2 , we see that T2 = T1



P2 P1

(n−1)/n (4.23)

and T2 = T1



V1 V2

n−1 (4.24)

Properties of Pure, Simple Compressible Substances

99

4.7.2 Work With P V n = constant, the work done will be  V2 dV P2 V2 − P1 V1 mR(T2 − T1 ) W=C = = n 1−n 1−n V1 V

(n = 1)

(4.25)

If n = 1, we see that the polytropic process reverts to the isothermal process with W = P1 V1 ln

V2 V2 = mRT1 ln V1 V1

(3.43)

4.7.3 Internal Energy and Enthalpy As long as the gas is an ideal gas with specific heats c p and c v assumed to be constant, the change in internal energy and enthalpy will always be U2 − U1 = m(u2 − u1 ) = mc v (T2 − T1 )

(4.26)

H2 − H1 = m(h 2 − h 1 ) = mc p (T2 − T1 )

(4.27)

and

4.7.4 Heat Transfer For any process involving a closed system with negligible change in kinetic and potential energy, the first law energy equation is Q = U + W For a polytropic process of an ideal gas beginning at state 1 and ending at state 2 with n = 1 Equations 4.25 and 4.26 give Q = U + W

mR(T2 − T1 ) 1−n   c v + R − nc v =m (T2 − T1 ) 1−n = mc v (T2 − T1 ) +

But c p = cv + R so that

 Q = mc v

c p = kc v

and

 k−n (T2 − T1 ) 1−n

or Q = mc n (T2 − T1 )

(4.28)

where c n is the polytropic specific heat  cn = cv

k−n 1−n

 (4.29)

100

Introduction to Thermal and Fluid Engineering P P=C

2

3

V=C Pv1.28

1 V V2

V1

FIGURE 4.18 P-v diagram for quasistatic processes of Example 4.10. Here V1 = 4V2 .

We note that for n = 0, which represents the constant pressure process cn = cv k = c p and if n = ±∞, which represents the constant volume process  cn = cv

k−n 1−n

⎛k

 = lim c v n−→∞

−1

⎞

n⎜n ⎟ ⎝ ⎠ = cv n 1 −1 n

If n = k, the process is an adiabatic process in which c n = 0 and Q = 0. The adiabatic process is the fifth process of the ideal gas that we will consider in Chapter 7.

Example 4.10 As illustrated in Figure 4.18, 8 g of air undergoes the following quasistatic processes in a piston-cylinder assembly: state 1 to state 2

a polytropic compression with n = 1.28 to one-quarter of the initial volume

state 2 to state 3

a constant pressure expansion to the initial volume

state 3 to state 1

a constant volume heat rejection

If the initial pressure is 5 bar and the initial temperature is 212◦ C (485 K), determine the net work done on the gas.

Solution Assumptions and Specifications (1) The air in the piston-cylinder assembly is a closed system. (2) The air obeys the ideal gas law. (3) Kinetic and potential energy changes are negligible. (4) Quasi-static processes are specified.

Properties of Pure, Simple Compressible Substances

101

The strategy here is to note that no work is done at constant volume from state 3 to state 1 and that W = W12 + W23 For air, Table A.1 gives R = 0.287 kJ/kg-K. Then at state 1 via the ideal gas law V1 = =

mRT1 P1 (0.008 kg)(287 N-m/kg-K)(485 K) 5 × 105 N/m2

= 2.227 × 10−3 m3 and with V2 = V1 /4 V2 =

2.227 × 10−3 m3 = 5.568 × 10−4 m3 4

we can obtain P2 from Equation 4.22  P2 = P1

V1 V2

n

= (5 × 105 N/m3 )(4) 1.28 = 2.949 × 106 N/m2 Because process 2-3 is a constant pressure process that returns the system to the original volume, we see that P3 = P2 and V3 = V1 P3 = 2.949 × 106 N/m2

and

V3 = 2.227 × 10−3 m3

All of the data required to determine the work done on the gas is at hand. From state 1 to state 2, Equation 4.25 gives W12 =

P2 V2 − P1 V1 1−n

With P2 V2 = (2.949 × 106 N/m2 )(5.568 × 10−4 m3 ) = 1642.0 N-m and P1 V1 = (0.500 × 106 N/m2 )(2.227 × 10−3 m3 ) = 1113.5 N-m we have W12 =

1642.0 N-m − 1113.5 N-m 528.5 N-m = = −1887.5 N-m 1 − 1.28 −0.28

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Introduction to Thermal and Fluid Engineering

From state 2 to state 3, n = 0 and Equation 4.25 gives W23 = P3 V3 − P2 V2 = P3 (V3 − V2 ) = (2.949 × 106 N/m2 )(2.227 × 10−3 m3 − 5.568 × 10−4 m3 ) = (2.949 × 106 N/m2 )(1.670 × 10−3 m3 ) = 4925.4 N-m The work done is W = W12 + W13 = −1887.5 N-m + 4925.4 N-m = + 3037.9 N-m ⇐ The positive sign indicates that the work is done by the air in the piston-cylinder assembly on the surroundings.

4.8

Summary

The state postulate supplies the number of variables that are required to completely specify the state of a single phase or homogeneous system. In the case of a simple, compressible, substance where Pdv is the only relevant work interaction, two independent properties are sufficient to completely specify the equilibrium state. The steam tables are categorized into •

Saturation tables

•

Superheated vapor tables

•

Compressed liquid tables

The refrigerant-134a tables are categorized into •

Saturation tables

•

Superheated vapor tables

Details for each of these tables have been provided. For an ideal gas, the equation of state is Pv = RT Several methods are available for determining the properties of a nonideal gas. They are •

The use of the method of corresponding states in which the compressibility factor Z=

Pv RT

is defined and charts are available to obtain values of Z. •

The use of equations of state proposed by various investigators.

Properties of Pure, Simple Compressible Substances

103

Two of the equations of state that are cited here are those of van der Waals 

P+

a  ¯ ( v¯ − b) = RT v¯ 2

and Redlich-Kwong P=

¯ RT a − 1/2 v¯ − b T v¯ ( v¯ + b)

Polytropic processes of an ideal gas are decribed by P Vn = C

(a constant)

where the exponent, n, can vary. Three processes of an ideal gas were considered in Chapter 3 and, in this chapter, the polytropic process, where n can take on any value, is considered. In particular, a polytropic specific heat is  cn = cv

k−n 1−n



where k = c p /c v .

4.9

Problems

Use of the Steam Tables 4.1: Given a water-steam system where T = 200◦ C with a quality of x = 0.18. Use the steam tables to determine the specific volume. 4.2: Given a water-steam system where P = 500 kPa with a quality of x = 0.36. Use the steam tables to determine the specific volume. 4.3: Given a water-steam system where v = 1.148 × 10−3 m3/kg. Use the steam tables to determine T, P, x, and h. 4.4: Given a water-steam system where v = 0.0502 m3/kg. Use the steam tables to determine T, P, x, and u. 4.5: Given a water-steam system where T = 300◦ C and x = 0.60. Use the steam tables to determine P, u, and s. 4.6: Given a water-steam system where P = 50 kPa and h = 1034.3 kJ/kg. Use the steam tables to determine T, x, and u. 4.7: Given a water-steam system where T = 200◦ C and P = 6 MPa. Use the steam tables to determine h and u. 4.8: Given a water-steam system where T = 400◦ C and P = 1.5 MPa. Use the steam tables to determine u, h, and s. 4.9: Given a water-steam system where P = 1 MPa and h = 2872.6 kJ/kg. Use the steam tables to determine T, u, and s. 4.10: Given a water-steam system where P = 1.25 MPa and T = 275◦ C. Use the steam tables to determine u.

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Systems Containing Water and/or Steam 4.11: Two steam boilers discharge equal amounts of steam into the same line. In boiler-1, the pressure is 800 kPa and the temperature is 600◦ C. In boiler-2, the pressure is 800 kPa and the quality is 92%. Determine (a) the enthalpy at the quilibrium condition and (b) the temperature at the equilibrium condition. 4.12: A 0.12-m3 drum contains a mixture of water and steam at 300◦ C. Determine (a) the volume occupied by each substance if their masses are equal and (b) the mass of each substance if their volumes are equal. 4.13: Heat is extracted from a saturated water-vapor mixture in a closed rigid container at 4 MPa. Determine the quality of the water-vapor mixture when it reaches 125◦ C. 4.14: Three kilograms of steam at 2 MPa and 400◦ C are expanded at constant volume to a pressure of 200 kPa. Determine (a) the volume of the container, (b) the final temperature, and (c) the final quality. 4.15: Determine the volume occupied by 0.525 kg of steam at 15 MPa and 88% quality. 4.16: One kilogram of wet steam at 200◦ C and with an enthalpy of 2250 kJ/kg is confined in a rigid container. Heat is applied until the steam becomes saturated. Determine (a) the initial pressure, (b) the initial quality, (c) the initial specific volume, and (d) the final pressure. 4.17: Two kilograms of steam at 1.5 MPa and a quality of x = 0.50 are expanded at constant pressure to an unknown final state. If 240 kJ of work are done by the steam, determine (a) the final temperature and (b) the heat added to the steam. 4.18: Two kilograms of steam at 1.6 MPa and a quality of x = 0.50 are expanded at constant pressure to an unknown final state. If 128 kJ of work are done by the steam, determine (a) the final temperature and (b) the heat added to the steam. 4.19: Five kilograms of steam at 2.5 MPa and 300◦ C undergo a constant pressure process until the quality becomes 50%. Determine (a) h, (b) u, (c) s, (d) W, and (e) Q. 4.20: State point 1 for a water-steam mixture inside of a weighted piston-cylinder assembly is at 3 MPa and 250◦ C. Heat is added until the temperature reaches state point 2 at 350◦ C. Determine (a) the work done to raise the piston and (b) the heat transferred. Use of Refrigerant R-134a Tables 4.21: Given a liquid-vapor refrigerant R-134a system where T = 48◦ C with a quality of x = 0.22. Determine the specific volume. 4.22: Given a liquid-vapor refrigerant R-134a system where P = 600 kPa with a quality of x = 0.42. Determine the specific volume. 4.23: Given a refrigerant R-134a system where v = 0.0409 m3 /kg. Determine T, P, x, and h. 4.24: Given a refrigerant R-134a system where v = 9.308 × 10−4 m3 /kg. Determine T, P, x, and u. 4.25: Given a liquid-vapor refrigerant R-134a system where T = 30◦ C and x = 0.40. Use the refrigerant R-134a tables to determine P, h, and s. 4.26: Given a refrigerant R-134a system where P = 8.6247 bars and h = 163 kJ/kg. Determine T, x, and u. 4.27: Given a refrigerant R-134a system where T = −10◦ C and P = 100 kPa. Determine u, h, and s.

Properties of Pure, Simple Compressible Substances

105

4.28: Given a refrigerant R-134a system where P = 400 kPa and h = 277.00 kJ/kg. Use the refrigerant R-134a tables to determine T, u, and s. 4.29: Refrigerant R-134a at 2 bars and 40◦ C is cooled at constant temperature until the final specific volume is 0.0165 m3 /kg. Determine the change in specific enthalpy. ˙ 4.30: A refrigerant R-134a system at 600 kPa and h = 120 kJ/kg is expanded at constant enthalpy to a pressure of 100 kPa. Determine the change in entropy. 4.31: Refrigerant R-134a is confined in a closed-rigid container having a volume of 0.125 m3 at 200 kPa and a quality of 0.488. Heat is added until the pressure reaches 400 kPa. Determine (a) the mass of refrigerant in the container and (b) the heat added to the refrigerant. 4.32: Refrigerant R-134a at 100 kPa and a specific volume of 0.160 m3 /kg undergoes a constant pressure nonflow process until the temperature is 0◦ C. For a mass of 1 kg, determine (a) h, (b) v, (c) s, (d) u, (e) W, and (f) Q. 4.33: Two kilograms of saturated refrigerant R-134a vapor at 100 kPa are compressed in an adiabatic nonflow process at constant entropy until the pressure is 1 MPa. Determine (a) the final temperature and (b) the work done. 4.34: Determine the heat rejected from 2.5 kg of refrigerant R-134a in a constant pressure process from 1.2 MPa and 60◦ C to a quality of 20%. 4.35: One half kilogram of refrigerant R-134a with a quality of x = 0.625 is expanded isothermally to a pressure of 600 kPa and 30◦ C. The measured work output during the expansion is 65 kJ/kg. Determine the heat transferred. Polytropic Processes of an Ideal Gas 4.36: Three kilograms of air are expanded polytropically (P V 1.28 = constant) from P1 = 600 kPa and T1 = 90◦ C to P2 = 200 kPa. Determine (a) V2 , (b) T2 , (c) H, (d) U, (e) W, and (f) Q. 4.37: Four kilograms of carbon dioxide are compressed polytropically (P V 1.25 = constant) from P1 = 100 kPa and T1 = 60◦ C to T2 = 227◦ C. Determine (a) P2 , (b) W, and (c) Q. 4.38: Nitrogen expands in a cylinder from 1500 kPa and 267◦ C to 100 kPa and 27◦ C. Determine the value of n for a polytropic (P V n = constant) process. 4.39: Two kilograms of air at 25◦ C are expanded in a polytropic process (P V 1.26 = constant) until the pressure is halved. Determine (a) U, (b) H , (c) W, and (d) Q. 4.40: Two kilograms of hydrogen undergo a polytropic compression from P1 = 6 MPa and v1 = 0.1875 m3 /kg until the pressure is doubled. Assume constant specific heats with the values given in Table A.1 and determine (a) v1 , (b) T1 and T2 , (c) H, (d) U, (e) W, and (f) Q. 4.41: In a polytropic process, hydrogen expands from P1 = 1 MPa, T1 = 60◦ C and V1 = 0.275 m3 to P2 = 100 kPa. The polytropic exponent is 1.3. Using the specific heats given in Table A.1 as constant, determine (a) the mass of hydrogen present, (b) T2 , (c) W, (d) Q, and (e) H. 4.42: Six kilograms of methane are compressed in a polytropic process (P V 1.18 = constant) from P1 = 100 kPa and T1 = 57◦ C to T2 = 257◦ C. Assuming that the methane is an ideal gas and that the specific heats listed in Table A.1 are constant, determine (a) P2 , (b) W, and (c) Q. 4.43: A quantity of propane expands polytropically (P V n = constant) from 1 MPa and 0.20 m3 to 10 MPa and 1.262 m3 . Determine (a) the value of n and (b) the work done.

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4.44: Two kilograms of helium undergoes a polytropic process (P V n = constant) from V1 = 4.092 m3 and P1 = 650 kPa to P2 = 100 kPa and T2 = 367◦ C. Determine (a) the value of n and (b) the work done. 4.45: One kilogram of a gas, which may be presumed to be ideal, has a value of R = 0.136 kJ/kg-K and k = 1.352. The gas undergoes a polytropic process (P V n = constant) from P1 = 100 kPa and T1 = 32◦ C to P2 = 500 kPa and V2 = 0.105 m3 . Determine (a) the value of n, (b) U, and (c) H. The Principle of Corresponding States 4.46: Determine the specific volume of superheated steam at 20 MPa and 500◦ C using (a) the steam tables, (b) the ideal gas equation, and (c) the principle of corresponding states. 4.47: Use the principle of corresponding states to obtain the density of oxygen at 4 MPa and −100◦ C. 4.48: Use the principle of corresponding states to determine the pressure at which the specific volume of oxygen at −100◦ C is 3.75 × 10−3 m3 /kg. 4.49: Propane at P1 = 2.5 MPa and v1 = 0.01875 m3 /kg is heated at constant volume until the pressure is P2 = 4 MPa. Use the principle of corresponding states to estimate the change in temperature. 4.50: Steam at 1 MPa and 327◦ C is heated at constant volume until the pressure is doubled. Use the principle of corresponding states to find the final temperature. 4.51: Steam at 10 MPa and 427◦ C is expanded isothermally until the volume is doubled. Use the principle of corresponding states to find the final pressure. 4.52: Steam at 8 MPa and 227◦ C is heated at constant pressure until the temperature is doubled. Use the principle of corresponding states to find the final specific volume. 4.53: Air at 1 MPa and 227◦ C flows through a 5.08 cm diameter tube at a velocity of Vˆ = 40 m/s. Determine the mass flow (a) by considering air to be an ideal gas and (b) by using the principle of corresponding states. 4.54: Consider refrigerant R-134a at 1.2 MPa and 63.6◦ C and determine its specific volume by (a) the ideal gas law and (b) the principle of corresponding states. 4.55: Refrigerant R-134a is heated at a constant pressure of 2 MPa from 82.3◦ C to 157.1◦ C. Use the principle of corresponding states to determine the work done. Equations of State 4.56: Two kilograms of propane at 20◦ C occupy a closed and rigid container having a volume of 0.05 m3 . Determine the pressure using (a) the ideal gas law, (b) the van der Waals equation of state, and (c) the Redlich-Kwong equation of state. 4.57: The specific volume of superheated steam at 1 MPa and 200◦ C is 0.2059 m3 /kg. Determine the specific volume of the steam using (a) the ideal gas law, (b) the principle of corresponding states, and (c) the van der Waals equation of state. Then use the steam table value as the basis for determining the percent error in each. 4.58: One kilogram of nitrogen at −110◦ C is contained in a closed and rigid vessel having a volume of 0.1 m3 . Determine the pressure exerted by the nitrogen using the van der Waals equation of state. 4.59: Use the Redlich-Kwong equation of state to find the pressure of carbon monoxide gas with a temperature of 310 K and a specific volume of 0.025 m3 /kg and compare the value with that predicted by the ideal gas equation.

Properties of Pure, Simple Compressible Substances

107

4.60: The pressure of R-134a vapor is 500 kPa at a temperature of 40◦ C. Determine the specific volume of the vapor by means of the van der Waals equation of state and compare the result with the value read from Table A.8. 4.61: Determine the compressibility factor for (a) Argon at 200 K and P = 5 MPa (b) Carbon dioxide at 500 K and P = 10 MPa (c) Methane at 382.4 K and P = 2.32 MPa (d) Helium at T = 15.9 K and P = 0.46 MPa 4.62: Oxygen gas in a container is at 210 K and has a specific volume of 0.002 m3/kg. Use the Redlich-Kwong equation of state to determine the pressure of the oxygen and then compare your result with the value obtained from the ideal gas equation. 4.63: For oxygen at 80 K with a specific volume of v = 0.0125 m3/kg, determine the pressure via (a) the ideal gas law, (b) the van der Waals equation of state, and (c) the RedlichKwong equation of state. 4.64: Hydrogen gas at 100 K has a specific volume of 1.98 m3/kg. Use the Beattie-Bridgeman equation of state P=

¯  RT c  A 1 − ( v¯ + B) − 2 2 3 v¯ v¯ T v¯

where  a A = Ao 1 − v¯

and

(kPa)

  b B = Bo 1 − v¯

with Ao = 20.0117

a = −0.00506

Bo = 0.02096

b = −0.04359

c = 0.0504 to find the pressure. All of the constants given will provide P in kPa as long as v¯ is in m3/kgmol.

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5 Control Volume Mass and Energy Analysis

Chapter Objectives •

To discuss what is meant by an open system, or a control volume.

•

To define what is meant by the conservation of mass with regard to a control volume.

To develop a relationship between the mass rate of flow and the volumetric rate of flow. • To discuss the conservation of energy and to show how energy equations, in accordance with the first law of thermodynamics, may be written for a control volume. • To develop the concept of flow work. •

To consider, in detail and on an energy and first law of thermodynamics basis, the analysis and performance of several of the devices that may make up a thermal system. • To give the definitions of synthesis and analysis and to provide a design example involving a “first law” heat balance. •

5.1

Introduction

Several mechanical “real-world” devices lend themselves to control volume analysis. Examples include nozzles, diffusers, turbines, compressors, pumps, mixing chambers, and heat exchangers. This chapter gives illustrations of control volume analysis as it pertains to these devices. We begin this development with a restatement of what is meant by closed and open systems. In Section 2.1, we defined a closed system or control mass as a system in which only energy (and not mass) may cross the system boundaries. An open system or control volume is a system in which both mass and energy can be transmitted across the system boundaries. The boundaries of a control volume are referred to as control surfaces. Flow through a control volume can be steady or transient. Steady flow implies no change with respect to time and transient flow occurs when the flow is not steady. The term uniform flow refers to flow that does not change with respect to location within a specified region.

109

110

5.2

Introduction to Thermal and Fluid Engineering

The Control Volume

As we have indicated, the difference between the control mass and the control volume formulation lies in the movement of mass across system boundaries. The mass transferred will contain energy. We will examine the conservation of mass and the conservation of energy for a control volume providing several illustrations. We turn next to a discussion of the conservation of mass (Section 5.3) and the conservation of energy (Section 5.4).

5.3

Conservation of Mass

5.3.1 The Mass Balance We considered the conservation of mass principle in Section 3.1 where we noted that Mass can neither be created nor destroyed and, except in chemical processes, its composition cannot be altered from one form to another. Thus, in the absence of nuclear reactions, mass is a conserved property, and the conservation of mass principle applied to the “multiple input-multiple output” control volume (designated by cv) shown in Figure 5.1 may be stated as ⎧ ⎪ ⎪ ⎨

⎫ the net change ⎪ ⎪ ⎬ in the mass within = a control volume ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ during a time period ⎧ ⎪ ⎪ ⎨

⎫ ⎧ ⎫ the summation of ⎪ the summation of ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ the mass entering a the mass leaving a − control volume during ⎪ ⎪ ⎪ control volume during ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ ⎩ ⎭ the time period the time period

. m3e3 . m1e1

. m2e2

FIGURE 5.1 A control volume with multiple inputs and outputs.

dm dt

. m4e4 . m5e5

Control Volume Mass and Energy Analysis

111 Control Surface

Vˆn

A

Δx

FIGURE 5.2 A fluid element of area A and length d x, located just upstream of a control volume.

On a rate basis, this may be written as ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ the rate of the rate of ⎨the rate of change⎬ ⎨ ⎬ ⎨ ⎬ of the mass within = mass entering mass leaving − ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ a control volume the control volume the control volume (5.1) or dmcv = m ˙i − m ˙e dt e i

(5.2)

where m ˙ is defined as the mass flow rate, mcv is the mass residing inside the control volume, and the subscripts, i and e, refer, respectively, to all inlets and exits. Because Equation 5.2 represents the conservation of mass, it is usually called the mass balance and is a form of what is referred to as a continuity relationship or merely as continuity. In the case of steady flow, where no mass accumulates within the control volume, dmcv /dt = 0 and Equation 5.2 reduces to m ˙i = m ˙e (5.3) i

e

and for a “single inlet-single outlet” control volume with state point 1 at the inlet and state point 2 at the outlet m ˙1 =m ˙2

(5.4)

5.3.2 The Volumetric Flow Rate The volumetric flow rate, typically in units of m3 /s, can be related to the mass flow rate, m, ˙ in kg/s, in terms of the local fluid density (or specific volume) and flow area. A fluid element that is located just “upstream” of a control surface is shown in Figure 5.2. With a cross-sectional area, A, we see that the volume of the fluid element is V = Ax where x is the distance that the fluid element traverses in time t. The average volumetric flow rate will be ˙ = lim Ax V t→0 t

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and because the normal velocity to the area d A is x Vˆ n = lim t→0 t we have ˙ = Vˆ n A V Uniform flow applies to a flow where all of the measurable fluid properties are uniform throughout any particular inlet or exit area. If the flow is not uniform over the area, dA, then we must integrate over the area

˙ = V Vˆ n dA A

and in this case, the mass flow rate will be

m ˙ = Vˆ n dA

(5.5)

A

If the flow is uniform over the entire cross-sectional area, Equation 5.5 reduces to m ˙ = AVˆ

(5.5a)

AVˆ v

(5.5b)

or, because v = 1/ m ˙ =

where Vˆ denotes the uniform value of the velocity at the inlet or outlet being considered. We call the product of the area, A, and the velocity Vˆ ˙ = AVˆ V

(5.6)

the volumetric flow rate.

Example 5.1 A pipe with an inner diameter of 30 cm carries 10,000 L/min of water and supplies a pipe tee with two branches. The smaller branch has a diameter of 10 cm and carries 2000 L/min and the larger branch has a diameter of 17.5 cm. Determine the velocities in each pipe.

Solution Assumptions and Specifications (1) The control volume involves the pipe tee with the contained fluid. (2) The water is incompressible. (3) Steady and uniform flow prevails. (4) There are no heat and work interactions with the environment. (5) The potential energy change is negligible. A sketch of the control volume with pertinent physical and flow data is shown in Figure 5.3. Note that the water flows are in liters per minute.

Control Volume Mass and Energy Analysis

113 17.5 cm

30 cm

10 cm

10,000 L/min

2000 L/min FIGURE 5.3 The control volume for Example 5.1.

The velocity at point 1 will be ˙1 V (10, 000 L/min)(10−3 m3 /L) Vˆ 1 = = = 141.5 m/min  A1 (0.30 m) 2 4 or Vˆ 1 = 2.36 m/s ⇐ At the pipe intersection continuity gives ˙1 = V ˙2 + V ˙3 V 10, 000 L/min = Vˆ 2 A2 + 2, 000 L/min The velocity at point 2 is therefore, (10, 000 L/min3 − 2000 L/min)(10−3 m3 /L) Vˆ 2 = = 332.6 m/min  (0.175 m) 2 4 or Vˆ 2 = 5.54 m/s ⇐ Finally, the velocity of the water at point 3 is (2000 L/min)(10−3 m3 /L) Vˆ 3 = = 254.7 m/min  (0.10 m) 2 4 or Vˆ 3 = 4.24 m/s ⇐

114

Introduction to Thermal and Fluid Engineering m1e1

Ecv(t)

z1

(a)

m2e2

Ecv(t + Δt)

z2

(b) FIGURE 5.4 Control volume showing a parcel of mass located (a) just “upstream” of the control volume entrance and (b) just “downstream” of the control volume exit.

5.4

Conservation of Energy for a Control Volume

5.4.1 Introduction We have noted that the significant difference between control volume and control mass analysis is that, in a control volume analysis, mass may flow across the system boundaries. Here, we will apply a form of the first law of thermodynamics to the control volume: Energy, a conservative property, can neither be created nor destroyed but can be changed or converted from one form to another. We will then see that, because of mass flow, m, ˙ we will need to replace the energy balance used in control mass analysis by an energy rate balance where the individual energy terms are in joules or kilojoules per second, which translate to watts or kilowatts. The control volumes to be employed in our development are shown in Figure 5.4. For each basic control volume, the surfaces are indicated by dashed lines. In addition, however, each of these control volumes shows a small mass. In Figure 5.4a, this mass is designated by the subscript 1 and its location is at the inlet to the control volume at state point i. In Figure 5.4b, the small mass is located at the outlet of the control volume at state point e and is designated by the subscript 2. This discussion is based on the fact that in the time interval, t, the entire mass m1 passes into the control volume and at the end of this time period, a quantity of mass, originally

Control Volume Mass and Energy Analysis

115

within the control volume has passed out to the exit region as indicated in Figure 5.4b. This allows us to identify the mass within the control volume with the subscript cv. We also specify that, consistent with the control mass analysis in Chapter 3, the conventions regarding the direction of the heat and work interactions through the control surfaces will be maintained. However, they will be energy rate quantities  ˙ cv −→ W and

 ˙ cv −→ Q

done by a system is positive done on a system is negative

transferred to a system is positive transferred from a system is negative

5.4.2 The Energy Rate Balance We begin our study of the energy rate balance with the conservation of energy principle for the control mass that occupies the entire interior of the control volume with a modification of Equation 3.12 E = E 2 − E 1 = Q − W

(3.12)

which, on a rate basis, can be represented as dE ˙ −W ˙ =Q dt Because this equation was written for a closed system, it cannot possibly account for any energy transfer due to the mass transport at the inlet and exit from the control volume. However, we can write equations that provide the energy associated with the control mass at both times, t + t and t E = E cv + E2 t+t

and

t+t

E = E cv + E 1 t

t

so that the change in energy within the control mass over the time interval, t, will be E − E = E cv − E cv + E 2 − E 1 t+t

t

t+t

t

But E 1 = m1 e 1

and

E 2 = m2 e 2

so that after we divide throughout by t we obtain E t+t − E t E cv,t+t − E cv,t m2 e 2 m1 e 1 = + − t t t t

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or in the limit as t −→ 0 dE d E cv = +m ˙ 2e2 − m ˙ 1e1 dt dt With this in Equation 3.12 we obtain d E cv ˙ cv − W ˙ +m ˙ e ee − m ˙ i ei = Q dt

(5.7)

where we have changed the subscripts so that m ˙1 →m ˙ i and m ˙2 →m ˙ e and we have added the subscript cv to the heat transmitted across the system boundaries. The reason for holding ˙ to W ˙ cv will be apparent in the next section. off on changing W At this point, we must make the distinction between work interactions at control surfaces where flow does and does not occur ˙ =W ˙ flow + W ˙ nonflow W where these terms are work rates (or power) and are expressed in watts or kilowatts. The ˙ nonflow term represents the contribution of the work interaction across the control volume W surfaces where no flow occurs. It is often referred to as Pdv work and we will refer to it as ˙ cv W ˙ nonflow = W ˙ Pd V = W ˙ cv W 5.4.3 Flow Work Because control volume analysis can (and does) involve mass flow across its boundaries, we need to recognize the effects of the work required to push a mass of fluid into or out of ˙ flow . the control volume. This work is called flow work and is represented as W In Figure 5.5 we see a portion of a control volume, its control surface and a single inlet located just “upstream” of the control volume. The inlet duct has a cross-sectional area, A, and, because of the pressure P that exists in the duct, there will be a force F = P A on the element of mass m. Because work is the product of force and distance, we have the flow work ˙ flow = PA dx dW and, in this case, the flow work rate will be

˙ flow = W P Ad x = P AVˆ A

Control Volume

F = PdA

dx FIGURE 5.5 A control volume, its control surface, and a single inlet located just upstream of the control volume.

Control Volume Mass and Energy Analysis

117

and because m ˙ =

AVˆ v

we have ˙ flow = m( W ˙ Pv)

(5.8)

For a single input-single output control volume with one-dimensional flow, we may write ˙ =W ˙ nonflow + W ˙ flow W as ˙ =W ˙ cv + m W ˙ i Pi vi − m ˙ e Pe ve

(5.9)

and we are ready, at last, to write the first law energy equation for the control volume. 5.4.4 The Control Volume Energy Equation The first law energy balance for the control volume will be ⎧ ⎫ ⎧ ⎫ the net rate of ⎪ the rate at which ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ change in the energy crosses the = energy within ⎪ ⎪ ⎪ control surfaces as ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ ⎩ ⎭ the control volume heat and/or work ⎧ ⎨

⎫ ⎧ ⎫ the rate at which the rate at which ⎬ ⎨ ⎬ + energy enters the control − energy exits the control ⎩ ⎭ ⎩ ⎭ volume by mass flow volume by mass flow With e i = ui +

Vˆ i2 + gzi 2

e e = ue +

Vˆ 2e + gze 2

and

this becomes, for a multiple input-multiple outlet system,

 ˆ2 V d E cv i ˙ cv − W ˙ cv + m ˙ i ui + Pi vi + =Q + gzi dt 2 i −

e

m ˙e

Vˆ 2 ue + Pe ve + e + gze 2

 (5.10)

The specific enthalpy was introduced in Section 3.8 h = u + Pv

(3.28b)

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Introduction to Thermal and Fluid Engineering

and may be employed here

 Vˆ i2 d E cv ˙ ˙ = Qcv − Wcv + m ˙ i hi + + gzi dt 2 i

 Vˆ 2 − m ˙ e h e + e + gze 2 e

(5.11)

If the flow is steady, dEcv /dt = 0, and Equation 5.11 becomes

 Vˆ i2 ˙ ˙ 0 = Qcv − Wcv + m ˙ i hi + + gzi 2 i

 Vˆ 2 − m ˙ e h e + e + gze 2 e

(5.12)

and for a single input-single output system in steady flow



 Vˆ i2 Vˆ 2e ˙ ˙ 0 = Qcv − Wcv + m ˙ hi + + gzi − m ˙ he + + gze 2 2

5.5

(5.13)

Specific Heats of Incompressible Substances

An incompressible substance is one whose specific volume is assumed to be constant and whose specific internal energy is assumed to vary only with temperature. Because the change in specific volume is zero, it can be shown that the constant volume and constant pressure specific heats for an incompressible substance are equal. Thus, c v and c p may be replaced by c without a subscript cv = c p = c

(5.14)

This indicates that the change in internal energy will be given by u2 − u1 = c(T2 − T1 )

(5.15)

and that the change in enthalpy will be h 2 − h 1 = u2 − u1 + v2 P2 − v1 P1 Moreover, with constant specific volume, v1 = v2 = v and h 2 − h 1 = u2 − u1 + v( P2 − P1 )

(5.16a)

h 2 − h 1 = c(T2 − T1 ) + v( P2 − P1 )

(5.16b)

or

Control Volume Mass and Energy Analysis

5.6

119

Applications of Control Volume Energy Analysis

5.6.1 Introduction Control volume energy analysis is fundamental in the estimation of the performance of equipment of all types. Seven types will be reviewed here and in each case, a fairly comprehensive example will be provided. The types of equipment include •

The nozzle and the diffuser

•

The turbine

•

The compressor

•

The pump

•

The open type of heat exchanger

•

The closed type of heat exchanger

•

The throttling valve

5.6.2 The Nozzle and the Diffuser A nozzle is a device that has a varying cross-sectional area that takes a fluid at high pressure and low velocity at its inlet and provides a high velocity stream at its outlet. The diffuser operates in exactly the opposite manner. It converts a high velocity low pressure fluid at the inlet to fluid at low velocity and a higher pressure at the outlet. We consider the nozzle in an example. Figure 5.6 shows an artist’s conception of both a nozzle and a diffuser. A combination of the two may be found in the sketch of the wind tunnel shown in Figure 5.7. The control volume analysis of the nozzle, with its control volume as indicated in Figure 5.8, is based on the assumptions that 1. The flow is steady. 2. The rate of heat transfer across the walls of the nozzle is negligible. 3. There is no work interaction between the system represented by the control volume and the environment. 4. Potential energy changes between inlet and outlet are negligible. V2 > V1 P2 < P1 1

FIGURE 5.6 (a) A nozzle and (b) a diffuser.

V2 < V1 P2 > P1 2

1

2

Nozzle

Diffuser

(a)

(b)

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Introduction to Thermal and Fluid Engineering Flow Straightening Screens Test Section

Diffuser Nozzle FIGURE 5.7 A wind tunnel may contain both a nozzle and diffuser.

With the foregoing assumptions in hand, we see that Equation 5.13 applies for this single input-single output system (m ˙1 =m ˙ 2 ) and that it may be written as h1 +

Vˆ 21 Vˆ 2 = h2 + 2 2 2

(5.17)

Example 5.2 A nozzle operates with steam. At the nozzle inlet (state point 1), P1 = 5 MPa,

T1 = 600◦ C and Vˆ 1 = 12 m/s and at the nozzle exit (state point 2), P2 = 1.5 MPa and Vˆ 2 = 800 m/s. If the mass flow rate through the nozzle is 2.25 kg/s, determine the nozzle exit area.

Solution Assumptions and Specifications (1) The system is the steam in the nozzle in steady flow. (2) The control volume is the nozzle. (3) There are no heat and/or work interactions between the system and the surroundings. (4) Potential energy changes are negligible. The control volume, with the specified conditions, is shown in Figure 5.9. The strategy is to find the nozzle outlet area from A2 =

m ˙ 2 v2 Vˆ 2 · m2 Vˆ2 P2 T2

· m 1 Vˆ1 P1 T1 2 1 FIGURE 5.8 The control volume for the nozzle.

· =m · m 1 2

Control Volume Mass and Energy Analysis

121 Control Volume Boundary · = 2.25 kg/s m

P1 = 5.0 MPa T1 = 600°C Vˆ1 = 12 m/s

P2 = 1.5 MPa Vˆ2 = 800 m/s 2 1 (a)

T 1

2

V (b) FIGURE 5.9 (a) The control volume for the nozzle in Example 5.2 and (b) the T-v diagram.

where the specific volume at the outlet, v2 , must be determined from P2 and some other independent intensive property. In this case, Equation 5.17 applies h1 +

Vˆ 21 Vˆ 2 = h2 + 2 2 2

and we may solve for h 2 , which is the other independent intensive property h2 = h1 +

Vˆ 21 − Vˆ 22 2

Table A.5 in Appendix A may be used to find h 1 at P1 = 5 MPa and T1 = 600◦ C h 1 = 3662.3 kJ/kg and with Vˆ 21 − Vˆ 22 (12 m/s) 2 − (800 m/s) 2 = 2 2 =

144 m2 /s2 − 6.40 × 105 m2 /s2 2

= −3.199 × 105 m2 /s2 = −319.93 kJ/kg

(5.17)

122

Introduction to Thermal and Fluid Engineering Rotating Blades Steam

Steam

Stationary Blades FIGURE 5.10 Cutaway view of an axial flow turbine.

the enthalpy at the outlet will be h2 = h1 +

Vˆ 21 − Vˆ 22 2

= 3662.3 kJ/kg − 319.93 kJ/kg = 3342.4 kJ/kg The two independent intensive properties needed to establish v2 are P2 and h 2 . Table A.5 with interpolation yields T2 = 440.6◦ C and v = 216.1 × 10−3 m3 /kg Thus, A2 =

mv ˙ 2 (2.25 kg/s)(216.1 × 10−3 m3 /kg) = = 6.078 × 10−4 m2 800 m/s Vˆ 2

or A2 = 6.078 cm2 ⇐ 5.6.3 The Turbine A reaction turbine is a device that possesses a set of blades that are mounted on a rotating shaft. Useful work is obtained from the working fluid as it expands between the inlet and outlet pressures. The gas turbine is treated in Chapter 8, the steam turbine is considered in Chapter 9, and impulse turbines are discussed in Chapters 14 and 18. A cutaway view of an axial flow turbine is shown in Figure 5.10. The example selected for this section is for a steam turbine. The control volume analysis of a turbine is based on the assumptions that 1. The flow is steady. 2. Potential energy changes between inlet and outlet are negligible. With the foregoing assumptions in hand, we see that Equation 5.13 applies for this single input-single output system and that it may be rewritten as



 Vˆ 22 Vˆ 21 ˙ cv = W ˙ cv + m Q ˙ h2 + −m ˙ h1 + (5.18) 2 2

Control Volume Mass and Energy Analysis

123

Example 5.3 Steam enters a turbine at a flow rate of 1.75 kg/s and develops a shaft power of 1020 kW. Conditions at the inlet are P1 = 80 bar, T1 = 400◦ C and Vˆ 1 = 12 m/s. At the turbine outlet, P2 = 10 kPa, Vˆ 2 = 48 m/s and the quality is 0.96 (96 %). The turbine operates at steady state and potential energy changes are negligible. Determine the heat loss through the turbine casing.

Solution Assumptions and Specifications (1) The control volume is the turbine. (2) The steam in the turbine is the system. (3) The system operates in the steady state. (4) Changes in potential energy are negligible. The configuration with the given data is shown in Figure 5.11 and in this case, Equation 5.18 applies



 Vˆ 22 Vˆ 21 ˙ cv = W ˙ cv + m Q ˙ h2 + −m ˙ h1 + 2 2 or



˙ cv = W ˙ cv + m Q ˙ (h 2 − h 1 ) +

Vˆ 22 − Vˆ 21 2



P1 = 80 bar T1 = 400°C Vˆ1 = 12 m/s

1

· W = 1020 kW · m = 1.75 kg/s P2 = 10 kPa Vˆ 2 = 48 m/s x = 0.96

2 (a) T 1

2 v (b) FIGURE 5.11 (a) The control volume for the turbine in Example 5.3 and (b) the T-v diagram.

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Introduction to Thermal and Fluid Engineering

The enthalpy at the inlet to the turbine is found from Table A.5 at 8 MPa and 400◦ C as h 1 = 3140.3 kJ/kg and at P2 = 10 kPa, we find from Table A.4 that h f = 191.7 kJ/kg

and

h fg = 2393.3 kJ/kg

Equation 4.6b then gives at state point 2 h 2 = h f + xh fg = 191.7 kJ/kg + (0.96)(2393.3 kJ/kg) = 191.7 kJ/kg + 2297.6 kJ/kg = 2489.3 kJ/kg Thus, h 2 − h 1 = 2489.3 kJ/kg − 3140.3 kJ/kg = −651.0 kJ/kg and Vˆ 22 − Vˆ 21 (48 m/s) 2 − (12 m/s) 2 = = 1080 m2 /s2 = 1.08 kJ/kg 2 2 Thus,  ˙ cv = W ˙ cv + m Q ˙ (h 2 − h 1 ) +

Vˆ 22 − Vˆ 21 2



= 1020 kW + (1.75 kg/s)(−651.0 kJ/kg + 1.08 kJ/kg) = 1020 kW + (1.75 kg/s)(−649.9 kJ/kg) = 1020 kW − 1137.4 kW = −117.4 kW ⇐ We note that this negative value means that heat is flowing from the turbine (the control volume) through the turbine casing out into the surrounding environment. We also note that the effect of the change in kinetic energy is much smaller than the effect of the specific enthalpy change.

5.6.4 The Compressor A compressor is a device whose purpose is to convert the work expended to drive it into an increase in pressure of the gas passing through it. Compressors may be of the reciprocating type, the axial flow type, or the centrifugal type. The control volume analysis of a compressor is based on the assumptions that 1. The flow is steady. 2. Kinetic and potential energy changes between inlet and outlet are negligible.

Control Volume Mass and Energy Analysis

125

With the foregoing assumptions in hand, we see that Equation 5.13 applies for this single input-single output system and that it may be rewritten as ˙ cv − W ˙ cv = m(h Q ˙ 2 − h1)

(5.19)

Example 5.4 Air enters a water-jacketed air compressor at P1 = 1 bar, T1 = 285 K and a velocity of Vˆ 1 = 6 m/s through an inlet area of A1 = 0.09375 m2 . At the exit, P2 = 8 bar, T2 = 480 K and the velocity is Vˆ 2 = 1.5 m/s. Heat enters the water jacket at the rate of 2.75 kW. Use the ideal gas model and determine the power required to drive the compressor.

Solution Assumptions and Specifications (1) The control volume is the compressor. (2) The air in the compressor is the system. (3) The flow is steady. (4) Kinetic and potential energy changes are negligible. (5) The air may be treated as an ideal gas. The configuration with the given data is shown in Figure 5.12. Here, the use of the ideal gas model is specified and we use Equation 5.19, which may be modified to give the work required to drive the compressor. ˙ cv = Q ˙ cv + m(h W ˙ 1 − h2) Because we are dealing with an ideal gas, the mass flow rate of the air can be found from the known conditions at the compressor inlet m ˙ =

A1 Vˆ 1 A1 Vˆ 1 P1 = v1 RT1

and with R and c p (which will be needed) obtained from Table A.1 R = 287 N-m/kg-K

and c p = 1.005 kJ/kg-K

we find m ˙ =

(0.09375 m2 )(6.0 m/s)(100 kPa) = 0.688 kg/s (287 N-m/kg-K)(285 K) . Q = 2.75 kW

P2 = 8 bar T2 = 480 K Vˆ 2 = 1.5 m/s

P1 = 1 bar T1 = 285 K Vˆ 1 = 6 m/s A = 0.09375 m2 1 FIGURE 5.12 Control volume for the compressor in Example 5.4.

2

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Introduction to Thermal and Fluid Engineering

The change in specific enthalpy will be h 1 − h 2 = c p (T1 − T2 ) = (1.005 kJ/kg-K)(285 K − 480 K) = (1.005 kJ/kg-K)(−195 K) = −195.8 kJ/kg ˙ cv is negative Because the heat flow is out of the control volume, Q ˙ cv = Q ˙ cv + m(h W ˙ 1 − h2) = −2.75 kW + (0.688 kg/s)(−195.8 kJ/kg) = −2.75 kW − 134.85 kJ/s = −137.6 kW ⇐ Our sign convention for work indicates it to be into the control volume and is, thus, the power required to drive the compressor. The horsepower required to drive the compressor will be HP = (137.6 kW)(1.34 HP/kW) = 184.4 HP ⇐

5.6.5 Pumps A pump is a device that changes the state of a liquid that passes through it as the working fluid. Usually, the work expended to drive the pump is converted to a change in head, either in the form of a change in pressure or a change in elevation. Pumps are treated in detail in Chapter 18. The control volume analysis of a pump is based on the assumptions that 1. The flow is steady. 2. The heat flow across the casing of the pump is negligible. With the foregoing assumptions in hand, we see that Equation 5.13 applies for this single input-single output system and that it may be rewritten as



 ˆ2 ˆ2 V V ˙ cv = m W ˙ i h i + i + gzi − m ˙ e h e + e + gze 2 2 or with m ˙i =m ˙e =m ˙ and the inlet and outlet designated by the subscripts 1 and 2 

  Vˆ 21 − Vˆ 22 ˙ cv = m W ˙ (h 1 − h 2 ) + + g(z1 − z2 ) 2

(5.20)

Example 5.5 Determine the required horsepower for a pump that carries water at a constant temperature and a flow rate of 8 kg/s through a piping system in which the outlet is 40 m higher than the inlet. The temperature in the pump and piping system is a constant 27◦ C. The pump conditions at the inlet are P1 = 1 bar and Vˆ 1 = 3 m/s and the conditions at the outlet are P2 = 2 bar and Vˆ 2 = 12 m/s. The pump is at a location where the acceleration of gravity may be taken as 9.81 m/s2 .

Control Volume Mass and Energy Analysis

127

Solution Assumptions and Specifications (1) The pump is the control volume. (2) The water at a constant temperature of 27◦ C is the system. (3) The system operates in steady flow. (4) There is no heat transfer to or from the surroundings. (5) The acceleration of gravity may be taken as 9.81 m/s2 . The configuration with the given data is shown in Figure 5.13 and we note that the acceleration of gravity at 9.81 m/s2 is specified. Under these conditions, Equation 5.20 is the energy rate balance 

  Vˆ 21 − Vˆ 22 ˙ cv = m W ˙ (h 1 − h 2 ) + + g(z1 − z2 ) (5.20) 2 We evaluate the kinetic and potential energy changes first. Vˆ 21 − Vˆ 22 (3 m/s) 2 − (12 m/s) 2 = = −67.5 m2 /s2 = −0.0675 kJ/kg 2 2 Then with z1 − z2 = −40 m g(z1 − z2 ) = (9.81 m/s2 )(−40 m) = −392.4 m2 /s2 = −0.392 kJ/kg Now     h 2 − h 1 = h f + v f ( P − Psat ) 2 − h f + v f ( P − Psat ) 1 and because T1 = T2 , the Psat and h f terms cancel, leaving h 2 − h 1 = v f ( P2 − P1 ) The value of v f is obtained from Table A.3 at T1 = T2 = 27◦ C (with interpolation). v f = 1.0034 × 10−3 m3 /kg

m· = 8 kg/s T = 27°C g = 9.81 m/s2 40 m

Pump

1

P1 = 1 bar ˆ = 3 m/s V 1

FIGURE 5.13 The control volume for the pump in Example 5.5.

P = 2 bar 2 Vˆ2 = 12 m/s 2

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Introduction to Thermal and Fluid Engineering

so that h 1 − h 2 = v f ( P1 − P2 ) = (1.0034 × 10−3 m3 /kg)(1 bar − 2 bar) = (1.0034 × 10−3 m3 /kg)(−105 N/m2 ) = −100.3 N-m/kg

− 0.1003 kJ/kg

or

Then 

˙ cv = m W ˙ (h 1 − h 2 ) +

Vˆ 21 − Vˆ 22 2



 + g(z1 − z2 )

and with a flow rate of 8 kg/s, we have ˙ cv = (8 kg/s)(−0.100 kJ/kg − 0.0675 kJ/kg − 0.392 kJ/kg) W = (8 kg/s)(−0.560 kJ/kg) = −4.48 kJ/s

or

− 4.48 kW

This negative value for the power interaction shows that work is done on the system. The horsepower required to drive the pump will be HP = (4.48 kW)(1.34 HP/kW) = 6.00 HP ⇐ 5.6.6 The Mixing Chamber A mixing chamber may have more than one inlet and outlet and because of this, conservation of mass requires a mass balance. For steady flow, where no mass accumulates inside the control volume, Equation 5.3 pertains

m ˙i =

i



m ˙e

(5.2)

e

After all mass flow rates are established we then write an energy rate balance based on 1. The flow is steady. 2. Heat flow across the control volume boundary is negligible. 3. There is no work interaction between the control volume and the surroundings. 4. All kinetic and potential energy changes are negligible. With these assumptions, we employ a modification of Equation 5.12, which applies to a multiple input-multiple output system under steady flow conditions i

m ˙ i hi =



m ˙ e he

(5.21)

e

Example 5.6 A water mixing chamber has two inlets and one outlet. At inlet 1, P1 = 8 bar, T1 = 200◦ C, and m ˙ 1 = 50 kg/s. At inlet 2, P2 = 8 bar, T2 = 60◦ C, and A2 = 22.5 cm2 . At the ˙ 3 = 0.075 m3 /s. Determine (a) the mass flow rate at the outlet, (b) outlet, P3 = 8 bar and V the mass flow rate at inlet 2, and (c) the velocity at inlet 2.

Control Volume Mass and Energy Analysis 1

129

P1 = 8 bar T1 = 200°C m· 1 = 50 kg/s

P2 = 8 bar 2 T2 = 60°C A2 = 22.5 cm2

3 P3 = 8 bar · V3 = 0.075 m3/s

FIGURE 5.14 The control volume for the mixing chamber in Example 5.6.

Solution Assumptions and Specifications (1) The mixing chamber is taken as the control volume. (2) The fluid within the chamber is taken as the system. (3) Operation is in steady flow. (4) There are negligible heat transfer and work interactions between the control volume and the surroundings. (5) There are negligible kinetic and potential energy changes. The configuration with the given data is shown in Figure 5.14. (a) The mass flow rate at the outlet can be obtained from a combination of Equations 5.5b and 5.6 m ˙3 =

˙3 V v3

where, v3 is the specific volume for the saturated liquid at 0.80 MPa and Table A.4 reveals that v3 = vf = 1.115 × 10−3 m3 /kg and m ˙3 =

˙3 V 0.075 m3 /s = = 67.27 kg/s ⇐ v3 1.115 × 10−3 m3 /kg

(b) For this control volume with two inlets and one outlet, we may use Equation 5.3 to obtain m ˙1+m ˙2 =m ˙3 so that m ˙2 =m ˙3−m ˙ 1 = 67.27 kg/s − 50 kg/s = 17.27 kg/s ⇐

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Introduction to Thermal and Fluid Engineering

(c) We find the velocity at inlet 2 from a modification of Equation 5.5b. m ˙ 2 v2 Vˆ 2 = A2 and because state point 2 is a compressed liquid v2 ≈ vf (T2 ) Table A.3 at 60◦ C gives vf = 1.017 × 10−3 m3 /kg so that m ˙ 2 v2 (17.27 kg/s)(1.017 × 10−3 m3 /kg) Vˆ 2 = = = 7.81 m/s ⇐ A2 22.5 × 10−4 m2

5.6.7 Heat Exchangers The primary purpose of a heat exchanger is to transfer energy between two fluids. They are treated in detail in Chapter 25. Here, we note that they are usually classified as regenerators, open-type exchangers, and closed-type exchangers or recuperators. Regenerators are heat exchangers in which the hot and cold fluids flow alternately through the same space with as little mixing between the two streams as possible. The quantity of energy transferred is dependent upon the fluid and flow properties of the fluid streams as well as the geometry and thermal properties of the surface. Open-type heat exchangers, as the name implies, are devices in which actual mixing of the two fluid streams occurs. The hot and cold fluids enter the open-type exchanger and leave as a single stream. We note that the mixing chamber considered in Section 5.6.6 and in Example 5.6 is an open-type heat exchanger. In the closed type of heat exchanger or recuperator, the fluid streams do not come into direct contact with one another and are separated by a tube wall or surface. The heat transfer is by convection from the hotter fluid to the surface, conduction through the tube wall or surface by conduction and then by convection to the cooler fluid. Typical examples of this type of heat exchanger are indicated in Figure 5.15. The energy rate balance for the open-type heat exchanger is much like that for the closed-type of exchanger or mixing chamber considered in Section 5.6.6. It is based on the following assumptions: 1. The flow is steady. 2. Because insulation can be employed to minimize the heat loss from the heat exchanger to the surroundings, the heat flow across the control volume boundary is assumed to be negligible. 3. Other than the energy carried by the fluid streams, there is no work interaction between the control volume and the surroundings. 4. All kinetic and potential energy changes are negligible.

Control Volume Mass and Energy Analysis

131

Fluid 2

Fluid 1 (a) Fluid 1

Fluid 2

(b) Shell-Side Fluid

Tube-Side Fluid

(c)

Hot Fluid

Cold Fluid (d)

FIGURE 5.15 Examples of closed-type heat exchangers or recuperators: (a) the counterflow double-pipe exchanger, (b) the cocurrent or parallel flow double-pipe exchanger, (c) the shell-and-tube heat exchanger with one shell pass and two tube passes, and (d) the compact cross-flow exchanger.

With these assumptions, we employ Equations 5.3 and 5.12 that apply to the multiple input-multiple output system. Under steady-state conditions, Equation 5.3 is the mass balance and Equation 5.12 is the energy rate balance, which can be modified to m ˙ i hi = m ˙ e he (5.22) i

e

While we observe that this is the same equation used for the mixing chamber, we note that, for a two-fluid exchanger, there are two inlets and two outlets.

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Introduction to Thermal and Fluid Engineering

Example 5.7 A steam power plant condenser takes steam at its inlet at 8 kPa and a quality of 0.96 (96%) and the condensate leaves as a saturated liquid at 8 kPa. Cooling water is supplied at atmospheric pressure by a river at 25◦ C and leaves at 35◦ C. The condensate flow rate is 2.75 kg/s and there is no pressure drop between the cooling water inlet and outlet. Determine (a) the ratio of the mass flow rate of the cooling water to the mass flow rate of the condensate, (b) the mass flow rate of the cooling water, and (c) the heat transferred from the condensate to the cooling water.

Solution Assumptions and Specifications (1) The overall condenser is divided into two control volumes. (2) The two flowing streams constitute the system. (3) Steady-state conditions apply. (4) There are no heat and/or work interactions between the overall control volume and the surroundings. (5) Kinetic and potential energy changes are negligible. (6) The cooling water is at constant pressure and its flow may be considered to be incompressible. Figure 5.16a shows the control volume for the entire steam condenser with the pertinent inlet and outlet conditions; Figure 5.16b is the control volume for just the steam side of the condenser. We note, in Figure 5.16b, that there is heat flow from the condensing steam to the cooling water but that this heat flow does not appear in the overall control volume shown in Figure 5.16a. We also note that the cooling water side is at constant pressure. (a) To find the ratio of the mass flow rate of the cooling water to the mass flow rate of the condensate, we first note that a mass balance for both the condensate and cooling water sides of the condenser will yield m ˙1 =m ˙2

and m ˙3 =m ˙4

Then, for the control volume in Figure 5.18a, Equation 5.22 gives m ˙ 1h1 + m ˙ 3h3 = m ˙ 2h2 + m ˙ 4h4 or m ˙ 1 (h 1 − h 2 ) = m ˙ 3 (h 4 − h 3 ) The mass flow rate ratio will be m ˙3 h1 − h2 = m ˙1 h4 − h3 and the problem becomes one of finding four enthalpy values. We can pinpoint h 2 as h f at P2 = 8 kPa and we find h f and h fg at 8 kPa from Table A.4 at Tsat,2 = 41.5◦ C as h 2 = h f = 173.8 kJ/kg

and

h fg = 2403.6 kJ/kg

Control Volume Mass and Energy Analysis

133

1

2

Steam 8 kPa x = 0.96 . m = 2.75 kg/s

Condensate 8 kPa

River Water 35°C

River Water 25°C

4

3 (a)

1

2

. Qcv (b) FIGURE 5.16 (a) The overall control volume for the entire steam condenser and (b) the control volume for the steam side of the condenser.

Then, use of Equation 4.6b gives h 1 h 1 = h f + xh fg = 173.8 kJ/kg + (0.96)(2403.6 kJ/kg) = 173.8 kJ/kg + 2307.5 kJ/kg = 2481.3 kJ/kg and h 1 − h 2 = 2481.3 kJ/kg − 173.8 kJ/kg = 2307.5 kJ/kg For the cooling water, the actual pressure is much greater than the saturation pressure at both T3 and T4 . This puts h 3 and h 4 into the compressed liquid region and, as indicated in Chapter 4, for compressed liquids at small temperature differences, h 4 − h 3 = h f,4 − h f,3

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Introduction to Thermal and Fluid Engineering

Thus, we read Table A.3 at T3 = 25◦ C and T4 = 35◦ C to obtain h 4 = 146.5 kJ/kg

and

h 3 = 104.8 kJ/kg

so that h 4 − h 3 = 146.5 kJ/kg − 104.8 kJ/kg = 41.7 kJ/kg Therefore, m ˙3 h1 − h2 = m ˙1 h4 − h3 =

2307.5 kJ/kg 41.7 kJ/kg

= 55.4 ⇐ (b) We establish the heat transferred between the condensing steam and the cooling water by employing the control volume of Figure 5.13b and we write the simple energy rate balance ˙ cv = m(h Q ˙ 2 − h1) Hence, ˙ cv = m(h Q ˙ 2 − h1) = (2.75 kg/s)(173.8 kJ/kg − 2481.3 kJ/kg) = (2.75 kg/s)(−2307.5, kJ/kg) = −6345.6 kJ/s

or

− 6.345 MW ⇐

The negative sign indicates that heat is flowing out of the control volume that contains the condensing steam. (c) The cooling water flow rate is merely m ˙ 3 = 55.4m ˙ 1 = (55.4)(2.75 kg/s) = 152.35 kg/s ⇐ 5.6.8 The Throttling Valve A throttling process is a steady flow process that occurs across a flow restriction with a subsequent reduction in the pressure of the flowing fluid. Throttling devices, as indicated in Figure 5.17, are usually in the form of a partially opened valve or a porous plug. Capillary tubes are also employed. A typical example of a throttling device is the throttling calorimeter (Figure 5.18), which measures the quality in a liquid vapor mixture. The analysis of the throttling device will be governed by the assumptions: 1. The flow is steady. 2. The heat flow across the control volume boundary is negligible. 3. There is no work interaction between the control volume and the surroundings. 4. All kinetic and potential energy changes are negligible.

Control Volume Mass and Energy Analysis

135

Exit

Inlet

Partially Open Valve (a) Exit

Inlet

Porous Plug (b) FIGURE 5.17 Two throttling devices.

With regard to Figure 5.18, these assumptions show that the mass balance of Equation 5.3 for this single input-single output system gives m ˙1 = m ˙2 = m ˙ and that the energy rate balance of Equation 5.13 reduces to m ˙ 1h1 = m ˙ 2h2 or because m ˙1 =m ˙2 h2 = h1

(5.23)

Thermometer

Calorimeter

Exhaust Valve FIGURE 5.18 Artist’s conception of the throttling calorimeter showing its control volume.

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Introduction to Thermal and Fluid Engineering

Example 5.8 Steam in the form of a liquid-vapor mixture at 2 MPa passes through a

throttling calorimeter (throttling valve) and is discharged at 1 bar and 150◦ C. Determine the quality of the steam in the supply line.

Solution Assumptions and Specifications (1) The control volume is the calorimeter shown in Figure 5.19. (2) The steam in the calorimeter is the system. (3) Steady-state conditions apply. (4) There are no heat and/or work interactions between the control volume and the surroundings. (5) Kinetic energy and potential energy changes are negligible. Figure 5.19 shows the control volume for the device and the foregoing assumptions and the fact that this is specified as a throttling valve allows us to use Equation 5.23 h1 = h2 Here, state point 2 is in the superheated region and we may read Table A.5 at 0.1 MPa and 150◦ C to obtain h 2 = 2776.8 kJ/kg Then, with h 1 = h 2 = 2776.8 kJ/kg, we use Table A.4 at 2 MPa to obtain h f = 908.3 kJ/kg

and

h fg = 1888.7 kJ/kg

and, from Equation 4.8, we find that x=

h1 − h f h fg

=

2776.8 kJ/kg − 908.3 kJ/kg 1888.7 kJ/kg

=

1868.5 kJ/kg 1888.7 kJ/kg

= 0.989

or

(98.9%) Temperature Measuring Device

. m1

P1

P1 = 2.0 MPa

P2

FIGURE 5.19 The control volume for the throttling calorimeter of Example 5.8.

. m2

P2 = 100 kPa

Control Volume Mass and Energy Analysis Input

137

System ?

Output

(a)

Input

System

Output ?

(b) FIGURE 5.20 Two systems. In (a) the input and output are both specified and the system is to be designed or synthesized. In (b) the system is specified and the output is to be found for a given input. This is an example of an analysis problem.

5.7

Synthesis or Analysis?

Synthesis refers to a procedure where a system is developed to achieve a desired result. On the other hand, analysis refers to the determination of the response of a specified system to a prescribed input. These may be cast into the system input-output framework of Figure 5.20 where synthesis and analysis are both described. As a somewhat trivial example, consider a system consisting of a single resistor shown in Figure 5.21. The input to this system is an applied voltage of 110 V and it is desired to obtain 5.5 A as the output. A simple synthesis procedure may be employed to obtain R=

V 110 V = = 20  I 5.5 A

which shows that the system consists of a single 20  resistor. If the resistance (system) is set at 40 , then the solution to the analysis problem for the same system input of 110 V will be I−

V 110 V = = 2.75 A R 40 

Almost all of the problems encountered in the thermal fluid sciences are not as simple as the foregoing “resistor” problem. In such cases, the system must be synthesized by analysis in which a system is selected and then subjected to analysis to determine whether the results are satisfactory. If not, the system is adjusted and the procedure is repeated until the specified result is achieved. This procedure may require several iterations and it is easy to see that the synthesis procedure may often be a monumental trial-and-error analysis problem.

110 V

55 A

(a)

R=?

110 V

I=?

??

(b)

FIGURE 5.21 The simple system illustrating (a) the synthesis problem and (b) the analysis problem.

138

Introduction to Thermal and Fluid Engineering 2

1

3

4

FIGURE 5.22 Control volume for a two-fluid heat exchanger.

5.8

The First Law Heat Balance

It was observed in Section 5.6.7 that the energy rate balance for a two-fluid heat exchanger (Figure 5.22) is given by Equation 5.22 ˙ = Q m ˙ i hi = m ˙ e he e

i

and for the case of a surface dissipating heat to a single fluid (Figure 5.23), this can be further modified to ˙ = m(h Q ˙ e − hi ) Almost every problem involving the flow of heat between two fluids or between a surface and a flowing fluid requires a first law energy rate balance or heat balance. Indeed, in Figure 5.22 for a liquid (m ˙ a ) flowing between two temperatures, T1 and T2 and another liquid (m ˙ b ) flowing between two temperatures, T3 and T4 , we have ˙ =m Q ˙ a (T1 − T2 ) = m ˙ b (T4 − T3 ) This is the heat balance showing that liquid a is the hotter fluid and liquid b is the cooler fluid. In Figure 5.23, with a single fluid flowing at a mass flow rate of m ˙ between two temper˙ will be atures T1 and T2 , the heat balance for a surface dissipation of Q ˙ = mc(T Q ˙ 1 − T2 ) A firm grasp on this fundamental problem, the so-called heat lost equals heat gained problem is required. Otherwise, the analyst will be shooting at a moving target and disappointment will be inevitable. . Q T2 . m T1 FIGURE 5.23 A surface dissipating heat to a single fluid.

Control Volume Mass and Energy Analysis

139

In Chapter 22, attention will be directed to the liquid cooling of an electronic component having a high heat dissipation. The determination of the component surface temperature requires an input from at least two of the disciplines treated in this book: •

It begins with a “first law” consideration in which the range of liquid flows to accommodate the specified heat dissipation are established.

•

The surface temperature is then found using procedures developed in Chapter 22.

Our attention now turns to the “first law” portion of the problem.

5.9

Design Example 1

A liquid cooled traveling wave tube collector is to be cooled in an application where the uniform dissipation in the collector is 1500 W. Coolanol-45 (Monsanto Chemical Co.), which enters the collector at 77◦ C, is to be used. The collector material is copper (k = 385 W/m-K) and in order to prevent “outgassing” of the copper that will destroy the vacuum, the walls of the collector cavity must be held to a maximum temperature of 150◦ C. The details of the collector cavity and the properties of coolanol-45 are provided in the second part of this design problem in Chapter 22. Coolanol-45 is but one of several liquids that are available and they all meet the requirements of high dielectric strength, chemical inertness, thermal decomposition and impurities, effects of moisture, pour and flash points, flammability, toxicity, surface tension, and vapor pressure. Note that water is not suitable because of its low dielectric strength.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The collector cavity walls are isothermal. (3) The heat flow is radially outward from the collector cavity to the coolant passage. (4) There is no heat flow longitudinally toward the ends of the collector. (5) Thermal properties vary with temperature in a prescribed manner. (6) The specific heat of the coolanol-45 varies in accordance with c = 3.40 × 10−3 T + 1.818 kJ/kg-K where T is in ◦ C. The calculations for the heat balance are displayed in Table 5.1. Observe that the liquid exit temperature is assumed and a specific heat is obtained. The coolant temperature increases are then calculated and the resulting exit temperature is then compared to the value assumed. It is almost always necessary to repeat the procedure until the exit temperatures match. This concludes the first law portion of the problem. The second portion of the problem involves forced convection heat transfer and the problem will be completed in Chapter 22.

140

Introduction to Thermal and Fluid Engineering Assume, m, ˙ kg/s T1 , ◦ C Assume T2 , ◦ C Tb , ◦ C c at Tb , J/kg-K ˙ mc, T = Q/ ˙ ◦C T2 , ◦ C T2 OK? Tb , ◦ C c at Tb , J/kg-K ˙ mc, T = Q/ ˙ ◦C T2 , ◦ C T2 OK?

5.10

0.100 77.0 84.6 80.8 2093 7.2 84.2 no 80.6 2092 7.2 84.2 yes

0.125 77.0 83.0 80.0 2090 5.7 82.7 no 79.9 2089 5.7 82.7 yes

0.150 77.0 81.8 79.4 2088 4.8 81.8 yes

0.175 77.0 81.0 79.0 2087 4.1 81.1 yes

0.200 77.0 80.6 78.8 2086 3.6 80.6 yes

Summary

A control volume represents an open system in which both mass and energy may be transferred across the system boundaries. By contrast, in Section 2.2.1, a closed system, often referred to as a control mass, was defined as a system in which energy may be transferred across its boundaries but there is no transfer of mass across the boundaries from or to the surroundings. ˙ is related to the fluid velocity and the area of The volumetric flow rate, designated as V, the flow cross section ˙ ˙ = AVˆ = m V v The conservation of energy principle leads to the energy rate balance for a control volume, which is

 ˆ2 V d E cv i ˙ cv − W ˙ cv + =Q m ˙ i hi + + gzi dt 2 i −

e

m ˙e

Vˆ 2 h e + e + gze 2

 (5.11)

For steady flow, this becomes ˙ cv − W ˙ cv + 0=Q

i

m ˙i

Vˆ 2 h i + i + gzi 2

 −



m ˙e

e

Vˆ 2 h e + e + gze 2



and for a single input-single output system, we have

Vˆ 2 ˙ cv − W ˙ cv + m 0=Q ˙ h i + i + gzi 2



Vˆ 2 −m ˙ h e + e + gze 2

 (5.13)

Control Volume Mass and Energy Analysis

141

For incompressible substances, no distinction is made between specific heats at constant volume and constant pressure. Thus, c p = c v = c and u2 − u1 = c(T2 − T1 )

h 2 − h 1 = c(T2 − T1 ) + v( P2 − P1 )

and

Control volume energy analysis is fundamental in the estimation of the performance of several types of equipment. •

For the control volume analysis of the nozzle and diffuser h1 +

•

•

For the turbine

Vˆ 21 Vˆ 2 = h2 + 2 2 2

(5.17)



 Vˆ 22 Vˆ 21 ˙ ˙ Qcv = Wcv + m ˙ h2 + −m ˙ h1 + 2 2

(5.18)

For the compressor ˙ cv − W ˙ cv = m(h Q ˙ 2 − h1)

•

For the pump 

˙ cv = m W ˙ (h 1 − h 2 ) + •

Vˆ 21 − Vˆ 22 2

m ˙ i hi =

+ g(z1 − z2 )

(5.20)



m ˙ e he

(5.21)

m ˙ e he

(5.22)

For the heat exchanger i

m ˙ i hi =

e

For the throttling valve h2 = h1

5.11



e

i

•



For the mixing chamber

•

(5.19)

(5.23)

Problems

Energy Equations 5.1: A steady-flow system has a mass flow rate of 4 kg/s of fluid entering the system at P1 = 0.6875 MPa, 1 = 3 kg/m3 , Vˆ 1 = 40 m/s, and u1 = 1600 kJ/kg. Conditions at the outlet of the system are P2 = 0.125 MPa, 2 = 0.75 kg/m3 , Vˆ 2 = 200 m/s, and u2 = 1480 kJ/kg. Each kilogram rejects 12 kJ of heat during its passage through the system. Determine the work done. 5.2: A steady-flow system takes in 0.75 kg/s of a fluid and discharges it at a point 27.5 m above the inlet. The fluid enters at a velocity of Vˆ 1 = 60 m/s and leaves at Vˆ 2 = 20 m/s.

142

Introduction to Thermal and Fluid Engineering During the process, 2 kW of heat are supplied from an external source and the increase in enthalpy is 1 kJ/kg. Determine the work done by the system.

5.3: A steam-generating unit receives 144,000 kg/h of water as saturated liquid at 280◦ C. The water leaves as saturated steam at 4 MPa. The heating value of the coal is 30,000 kJ/kg and air to support the combustion is supplied in the ratio of 13.5 kg of air/kg of coal. The air enters with an enthalpy of 50 kJ/kg and leaves with an enthalpy of 300 kJ/kg. Determine the amount of coal supplied. 5.4: A cylindrical tank with a diameter of 1.50 m has a hole in the bottom that has a diameter of 1.25 cm; a steady stream of water flows from the hole. Determine the quantity of ˙ that must be added to the contents of the tank to maintain the water level at flow, V, 2.5 m. 5.5: Water at 10 bar (gage) and 25◦ C enters a hydraulic turbine with a volumetric flow rate of 40 m3 /s at a velocity of 1.25 m/s at a height of 125 m above a given datum. The water leaves the turbine at 2 bar (gage) and 25.2◦ C at the datum level at a velocity of 10 m/s. Assume that the water is incompressible and that the turbine loses 4 J/kg of water flowing through it and determine (a) u, (b)  ke, (c)  pe, (d) h, ˙ cv . and (e) W 5.6: Assuming that water is incompressible and that no heat is transferred, determine the power required to pump 20 kg/s of water at 101 kPa and 75◦ C from a tank whose outlet is 5 m below ground level if the pipe diameter is 10 cm. 5.7: Refrigerant R-134a flows through a horizontal pipe reducer with the following inlet (point 1) and outlet (point 2) conditions: Inlet 10 40 6.25 7.52

Outlet P, bar T ◦C ˆ m/s V,

5 70

d, cm

5.08

Determine (a) the mass flow and (b) the exit velocity. 5.8: An insulated container is large enough to contain 4 kg of water and 1 kg of ice at 0◦ C with heat of fusion of 333 kJ/kg. A copper sphere having a mass of 16 kg at 98◦ C is dropped without splashing into the water-ice mixture. Determine the equilibrium temperature of the water-ice-copper system. 5.9: Determine how long it will take to heat a mountain cabin measuring 8 m × 4 m × 3 m containing air at 0◦ C to a comfort level of 21◦ C using a 2-kW heater. 5.10: A compressible fluid is flowing in a device with a single input and a single output. The data for the fluid and the device are as follows: Inlet (point 1) 6.8 6.25 58 290 0.720 0.280 12.70 0

Outlet (point 2) P, bar abs , kg/m3 ˆ m/s V, u, kJ/kg c v , kJ/kg-K R, kJ/kg-K d, cm z, m

4.0 0.80 680 125 0.720 0.280 10.37 4

Control Volume Mass and Energy Analysis

143

Determine between inlet and outlet (a)  ke, (b) h, (c)  pe, (d) the mass flow rate, (e) the total heat transferred if no work is done between inlet and outlet, and (f) the specific heat at constant pressure if an ideal gas is assumed. 5.11: Water flows at the rate of 750 L/min through the venturi shown in Figure P5.11. ˙ cv = W ˙ cv = u = 0 and determine the pressure drop between points Assume that Q 1 and 2. 1 2

6 cm 750 L/m 2 cm FIGURE P5.11

5.12: A compressor takes in 30 m3 /min of air having a density of 1.266 kg/m3 . The device discharges the air with a density of 4.870 kg/m3 . At the inlet, P1 = 100 kPa and at the outlet, P2 = 550 kPa. The increase in internal energy is 15 kJ/kg and the heat rejected by the air in passing through the compressor is 6 kJ/kg. Neglecting changes in kinetic and potential energy, determine the work required to drive the compressor. Nozzles and Diffusers 5.13: Steam at 20 bar and 800◦ C at a velocity of 75 m/s is provided to the inlet (point 1) of a nozzle. Determine the exit velocity (point 2) if the exit conditions are a liquid-vapor mixture at 5 bar and 92% quality. 5.14: Air at 4 bar and 67◦ C enters a nozzle with a mass flow rate of 0.625 kg/s. The exit area is 0.0025 m2 and the exit pressure and velocity are 1.20 bar and 240 m/s, respectively. Determine (a) the exit temperature assuming that the air is an ideal gas and that constant specific heats may be used, (b) the change in specific kinetic energy, and (c) the inlet velocity. 5.15: Steam at 4 MPa and 400◦ C at a mass flow rate of 0.75 kg/s is introduced at negligible inlet velocity to a horizontal nozzle. The steam exits the nozzle at 1.5 MPa at a velocity of 750 m/s. Determine (a) the exit condition (temperature or quality) and (b) the exit area. 5.16: Air, which may be treated as an ideal gas, enters a converging nozzle at 400 kPa and 247◦ C with a velocity of 40 m/s through an inlet area of 125 cm2 . The air leaves the nozzle at 100 kPa and a velocity of 200 m/s. During its passage through the nozzle, the air loses 40 kJ/kg of air flow. Determine (a) the mass flow rate of the air, (b) its outlet temperature, and (c) the inlet area of the nozzle. 5.17: Steam enters a horizontal and adiabatic nozzle at 4 MPa and 600◦ C at a velocity of 240 m/s through an area of 60 cm2 . The steam leaves the nozzle at 1.5 MPa and 300◦ C. Determine (a) the mass flow rate of the steam, (b) the exit velocity of the steam, and (c) the exit area of the nozzle. 5.18: Refrigerant R-134a at a steady flow rate of 2 kg/s enters an adiabatic nozzle at 500 kPa and 100◦ C with a velocity of 50 m/s. The fluid leaves the nozzle at 200 kPa and a velocity of 320 m/s. Determine (a) the exit temperature and (b) the inlet/outlet area ratio.

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Introduction to Thermal and Fluid Engineering

5.19: Air in steady flow enters an adiabatic diffuser through an inlet flow area of 100 cm2 at 100 kPa, 147◦ C, and 250 m/s and leaves at 150 kPa and 20 m/s. Consider the air as an ideal gas and determine (a) the exit temperature and (b) the exit area. 5.20: Saturated steam at 250◦ C at a steady flow rate enters a diffuser at a velocity of 600 m/s through an inlet flow area of 125 cm2 . The steam is discharged at 2.5 MPa and 280◦ C. Determine (a) the mass flow rate through the diffuser, (b) the exit velocity, and (c) the exit area. 5.21: Water, which may be taken as incompressible, enters an adiabatic diffuser in steady flow at 1 bar, 25◦ C, and a velocity of 90 m/s through an inlet flow area of 250 cm2 . The water leaves the diffuser at 5 bar and an exit area of 1500 cm2 . Determine (a) the exit velocity and (b) the outlet temperature. 5.22: Steam enters a diffuser as saturated vapor at 125◦ C with a velocity of 275 m/s. The exit pressure and temperature are 300 kPa and 150◦ C; and the exit area is 62.5 cm2 . Determine (a) exit velocity and (b) the mass flow rate. Turbines 5.23: Steam at a mass flow rate of 15 kg/s at 10 MPa and 500◦ C enters a well-insulated turbine at a velocity of 100 m/s. At the exit, the steam is at 1 MPa and 96% quality with a velocity of 40 m/s. Determine (a) the power output and (b) the inlet area. 5.24: An adiabatic turbine delivers 10 MW using steam at a steady-flow rate of 20 kg/s. Inlet conditions are 10 MPa and 500◦ C and the outlet pressure is 1 MPa. Potential and kinetic energy changes are negligible. Determine the condition of the outlet steam by giving either a temperature or a quality. 5.25: An adiabatic turbine uses steam to deliver 10 MW at a steady-flow rate of 10 kg/s. Inlet conditions are 10 MPa and 500◦ C and the outlet pressure is 1 MPa. Potential and kinetic energy changes are negligible. Determine the condition of the outlet steam by giving either a temperature or a quality. 5.26: Nitrogen, which may be taken as an ideal gas, expands in an adiabatic turbine from 1 MPa and 475◦ C to a pressure of 200 kPa. Inlet and exit velocities are 100 m/s and 180 m/s, respectively. The power delivered by the turbine is 280 kW. The inlet area is 62.5 cm2 . Determine the outlet temperature. 5.27: A steam turbine is designed to have a power output of 10 MW when the steady-flow mass flow rate is 16 kg/s and the steam is supplied at 4 MPa and 375◦ C, at a velocity of 200 m/s. The steam is discharged in a dry saturated condition at 20 kPa at a velocity of 60 m/s. Determine the heat transfer through the turbine casing. 5.28: Determine the mass flow rate of steam required to produce 750 kW output in a turbine where potential and kinetic energy changes are negligible. The steam enters at 1 MPa and 250◦ C and leaves at 200 kPa and a quality of 92%. 5.29: A steam turbine operates with inlet conditions of 10 MPa, 500◦ C, and a velocity of 80 m/s. Dry saturated steam leaves the outlet at 0.1 MPa at a velocity of 160 m/s. The datum level is taken at the outlet and the elevation of the inlet is at 2 m. The heat loss through the turbine casing is 40 kW and the mass flow rate is 2 kg/s. Determine (a) the power delivered by the turbine and (b) the power delivered by the turbine if potential and kinetic energy changes are negligible. 5.30: Determine the mass flow rate of steam required to produce 750 kW output in a turbine where potential and kinetic energy changes are negligible. The steam enters at 1 MPa and 250◦ C and leaves at 200 kPa and a quality of 92%.

Control Volume Mass and Energy Analysis

145

5.31: Air, which may be taken as an ideal gas, enters a turbine in a steady-flow process at 750 kPa and 487◦ C and 140 m/s. The air leaves the turbine at 150 kPa, 187◦ C, and 240 m/s. The inlet area is 5.625 cm2 and during passage through the turbine, the air loses 12.5 kJ/kg per kg/s of flow through the casing. Determine the power output. 5.32: Refrigerant R-134a enters a steady-flow turbine at 800 kPa, 80◦ C, and 60 m/s. Exit conditions are 0.60 bar, −10◦ C, and 120 m/s. The inlet area is 27.5 cm2 and the power developed is 280 kW. Determine (a) the mass flow rate and (b) the heat transferred through the turbine casing. Compressors and Pumps 5.33: Air enters a compressor at 100 kPa and 20◦ C with a velocity of 2 m/s and leaves at 600 kPa. The power input to the compressor is 500 kW of which 120 kW is lost to the surroundings in the form of heat. The area at the compressor inlet is 0.25 m2 . Assuming that the air behaves as an ideal gas and neglecting kinetic and potential energy changes, determine the temperature of the compressed air leaving the compressor. 5.34: In a vapor compression refrigerator, refrigerant R-134a enters the compressor as a saturated vapor at −4◦ C. The refrigerant leaves the compressor as a superheated vapor at 1 MPa and 50◦ C and the mass flow of the refrigerant is 0.02 kg/s. The compressor requires a work input of 750 W and kinetic and potential energy changes are negligible. Determine the heat lost to the surroundings. 5.35: In a water-cooled air compressor, the inlet pressure and temperature are 100 kPa and 298 K, respectively. Air leaves the compressor at a pressure of 500 kPa. The compression process follows the law Pv1.3 = constant and heat rejected to the cooling water amounts to 35 kW/kg of air. Assuming a mass flow of 2 kg/s and neglecting kinetic and potential energy change, determine the power input to the compressor. 5.36: An adiabatic compressor draws carbon dioxide gas at 95 kPa and 298 K and delivers it at 300 kPa and 450 K. The volumetric flow rate of carbon dioxide is 0.6 m3 /kg. Kinetic and potential energy changes during the compression process are negligible and the carbon dioxide can be treated as an ideal gas. For a flow rate of 2 kg/s, determine (a) the volumetric flow rate at the exit of the compressor and (b) the power required to drive the compressor. 5.37: Air enters a compressor with a mass flow rate of 1.25 kg/s at 105 kPa and 27◦ C and exits at 675 kPa and a specific volume of 0.1575 m3 /kg. The work required to drive the compressor is 200 kW. The heat lost in this process is absorbed by an insulated water jacket where the temperature rises to 294 K from an initial value of 282 K with no change in pressure. Any heat loss from the water jacket may be ignored and all kinetic and potential energy changes are negligible. Determine the mass flow of the cooling water in the water jacket. 5.38: A steady-state compression of nitrogen takes place in a compressor between 1 bar and 25◦ C at a mass flow rate of 0.875 kg/s and a velocity of 20 m/s to 2.5 bar and 375◦ C at a velocity of 100 m/s. Potential energy change and heat exchange with the environment are negligible. Determine the power required to drive the compressor. 5.39: The feedwater pump in a steam power plant takes in saturated liquid water at 20 kPa with a velocity of 1.5 m/s and delivers it at 1 MPa with a velocity of 10 m/s. The difference in elevation between the inlet and the outlet of the pump is 15 m and the water temperature remains constant at 60.1◦ C. The acceleration of gravity is taken as 9.81 m/s2 and the water is to be delivered at 25 kg/s. Determine the power required to operate the pump.

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5.40: In a 200-horsepower pump installed in a long pipeline the increase in kinetic energy of the oil is 8.4 kJ/s-kg but the potential energy change is negligible. If the temperature of the oil increases by 8◦ C and the oil may be considered as incompressible with a density of 860 kg/m3 with a specific heat of 1.909 kJ/kg-K. The difference in pressure between the inlet and the outlet is 12 kPa. Determine the mass flow rate of the oil. 5.41: A pump is used to deliver water from an underground tank to an overhead tank at a rate of 20 kg/s. The flow velocities in the inlet and outlet pipes are 1 m/s and 5 m/s, respectively. Assuming no change in water pressure and the water temperature constant at 300 K, determine the pumping power required if the water experiences a change in elevation of 15 m. 5.42: Water is delivered in a steady flow by a pump at a volumetric flow rate of 6.25 × 10−3 m3 /s through a pipe having an inner diameter (ID) of 18 cm. The outlet pipe is 80 m above the inlet and has an ID of 16 cm. Both the inlet and outlet pressures (1.25 bar) and temperatures (27◦ C) are nearly equal. The acceleration of gravity is 9.81 m/s2 . Determine the power required. Mixing Chambers 5.43: Two streams of steam enter a mixing chamber and exit as a single stream. Stream 1 has P1 = 500 kPa, x1 = 0.50, and m ˙ 1 = 2 kg/s. For stream 2, P1 = 500 kPa, T2 = 350◦ C, and m ˙ 2 = 4 kg/s and for the exit (stream 3), P3 = 500 kPa, and m ˙ 3 = 6 kg/s. Assuming that the mixing chamber is adiabatic, determine the enthalpy, quality, and the temperature of the outlet stream. 5.44: Water at 200 kPa and 110◦ C flows into a mixing device at the rate of 5 kg/s. The water is heated by mixing it with superheated steam at 200 kPa and 500◦ C. The outlet stream consists of saturated liquid water at 200 kPa. Assume that there is no heat loss to the surroundings and determine the mass flow of the superheated steam. 5.45: Two streams of refrigerant R-134a enter a mixing chamber and exit as a single stream. Stream 1 is saturated liquid with T1 = 8◦ C and m ˙ 1 = 1.5 kg/s. Stream 2 is saturated vapor with T2 = 8◦ C and m ˙ 2 = 2.2 kg/s. Heat is gained by the mixing chamber at the rate of 20 kW. Determine (a) the quality of the outlet stream and (b) the outlet area if the velocity at the outlet is 3 m/s. 5.46: In an adiabatic mixing chamber, two streams of refrigerant R-134a mix to form a single stream. Stream 1 is compressed liquid with P1 = 0.20 MPa, T1 = −20◦ C, and m ˙ 1 = 0.6 kg/s while stream 2 is superheated vapor with P2 = 0.20 MPa and T2 = 30◦ C. Determine the mass flow of stream 2 if the outlet tream is to have a pressure of 0.20 MPa and a quality of 0.75. 5.47: The pressure in an adiabatic mixing chamber is 1 MPa. Steam enters at inlet 1 at a rate of 3.0 kg/s at a quality of 90%. Steam enters inlet 2 at 4.5 kg/s and 300◦ C. The exit velocity (at point 3) is 9 m/s. Determine (a) the exit temperature and (b) the diameter of the exit pipe. 5.48: Hot water (inlet 1) enters an adiabatic mixing chamber at 300 kPa and 75◦ C at a flow rate of 0.625 kg/s. It is mixed with a stream of cold water (inlet 2) at 300 kPa and 15◦ C. If the mixture is to leave the chamber at 300 kPa and 50◦ C, determine the flow rate of the cold water stream. 5.49: A hot stream of refrigerant R-134a enters an adiabatic mixing chamber at 8 bar and 80◦ C. A cold stream enters the chamber at 8 bar as a saturated liquid. The ratio of the mass flow rates of the hot to cold streams is 2.75:1. Determine the condition of the

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147

stream leaving at 8 bar (either a temperature if superheated vapor or a quality if a liquid-vapor mixture). 5.50: At point 1, refrigerant R-134a enters an adiabatic mixing chamber as saturated vapor at 6 bar. At point 2 liquid enters at 6 bar as saturated liquid. The flow rates of each stream are identical. Determine the quality of the exiting stream. Heat Exchangers 5.51: In an air-water heat exchanger, air enters the heat exchanger at 125 kPa and 60◦ C and leaves at 37◦ C at a flow rate of 2 kg/s. The water enters at 20◦ C with a mass flow rate of 0.4 kg/s, the heat loss to the surroundings is estimated to be 5 kW and the air may be assumed to be an ideal gas. Determine the outlet temperature of the water. 5.52: Water at 1 MPa enters a boiler as saturated liquid and leaves as saturated vapor at 0.8 MPa. The stream enters a superheater where its temperature is raised to 400◦ C without further pressure loss. For a steam mass flow rate of 48 kg/s, determine the rate of heat supplied in (a) the boiler and (b) the superheater. 5.53: The condenser of a refrigerator receives R-134a at 400 kPa and 20◦ C and cools it to saturated liquid at 360 kPa. Determine the heat rejected in the condenser per kilogram of refrigerant. 5.54: In a cross-flow heat exchanger, air enters at 27◦ C and flows over a bank of tubes with a mass flow rate of 0.7 kg/s. The air is heated by the products of combustion flowing inside the tubes, which enter at a rate of 0.30 kg/s at 300◦ C and leave at 260◦ C. Both the air and the products of combustion with c p = 1.041 kJ/kg-K may be assumed to behave like ideal gases. Determine the temperature of the air leaving the heat exchanger. 5.55: In a steam power plant, steam, flowing at the rate of 160,000 kg/h, enters the condenser as a mixture at 20 kPa with a quality of 0.92 and is condensed into saturated liquid with no pressure loss. Cooling water drawn from a river enters the condenser tubes at 20◦ C. If the maximum temperature rise of the cooling water is not to exceed 8◦ C, determine the mass flow rate of water circulating through the condenser tubes. 5.56: Refrigerant R-134a is contained in the cooling unit of an air-conditioning system. It enters at a volumetric flow rate of 36 m3 /s at 1.15 bar and 40◦ C and exits at 1.12 bar and 20◦ C. The refrigerant enters the unit at 8◦ C with a quality of 36% and leaves as saturated vapor at 8◦ C. Neglecting kinetic and potential energy changes and any heat transfer across the enclosure of the cooling unit, determine (a) the heat transferred between the air and the refrigerant and (b) the mass flow rate of the refrigerant. 5.57: In an air preheater, flue gases from a boiler are used to preheat the incoming combustion air. The flue gases, at a flow rate of 1.8 kg/s, enter at a pressure of 102 kPa and a temperature of 525◦ C. Outside air at a mass flow rate of 1.7 m/s enters at 20◦ C and leaves at 265◦ C. The flue gases may be treated an ideal gas having the properties of air. Neglect kinetic and potential energy changes and any heat transfer across the enclosure of the preheater. Determine (a) the heat transferred in the air heater and (b) the exiting flue gas temperature. 5.58: In a steady-flow heat exchanger, steam enters at 200 kPa and 300◦ C and leaves at 175 kPa with a quality of 72%. This cooling is accomplished by 48 kg/min of air entering at 32◦ C and leaving at 48◦ C. Neglecting kinetic and potential energy changes and any heat transfer across the enclosure of the exchanger, determine the mass flow of the steam in kg/min.

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5.59: In a heat exchanger, air is heated by passing it over tubes containing steam which enters the tubes at 300 kPa and 250◦ C and leaves at 200 kPa and 200◦ C at a flow rate of 0.15 kg/s. The air enters at 101.3 kPa and 17◦ C and leaves at 37◦ C. The tubing has an inside diameter of 2 cm and the steam velocity is 4.8 m/s. Neglecting kinetic and potential energy changes and any heat transfer across the boundaries of the enclosure, determine (a) the heat transferred to the air, (b) the mass flow rate of the air, and (c) the number of tubes in the exchanger. 5.60: In the evaporator coil of an air-conditioning machine, dry air, which may be taken as an ideal gas at 30◦ C and atmospheric pressure, enters at a mass flow rate of 1.25 m/s. Refrigerant R-134a at −12◦ C and a quality of 22% is used in the coil. The refrigerant leaves the coil as saturated vapor and 24 kW are absorbed by the refrigerant. Assuming that both the heating and cooling processes take place at constant pressure with negligible kinetic and potential energy changes and no heat transfer across the boundaries of the system, determine (a) the flow rate of the refrigerant and (b) the exit temperature of the air. Throttling Devices 5.61: Saturated liquid water at 80◦ C passes through a throttling valve where its pressure drops to 35 kPa and the water becomes a liquid-vapor mixture. For the liquid-vapor mixture, determine (a) the temperature, (b) the quality, (c) the specific volume, and (d) the specific internal energy. 5.62: The properties of R-134a after throttling are P2 = 60 kPa and x2 = 0.32. Assuming the refrigerant at the inlet to the throttling device is in the saturated liquid state, determine the pressure and temperature before throttling. 5.63: A porous plug is placed in a steam supply line carrying saturated vapor at 1 MPa. As the steam flows through the porous plug, its pressure drops to 0.2 MPa. Assuming that the flow through the plug is an adiabatic throttling process, determine the temperature of the steam after its passage through the porous plug. 5.64: At the inlet of a throttling valve, the state of refrigerant R-134a is P1 = 700 kPa, and T1 = 20◦ C. At the outlet, the refrigerant is a liquid-vapor mixture at 100 kPa. Assume the process to be adiabatic with negligible changes in kinetic and potential energies and determine at the outlet, (a) the quality, (b) the specific volume, and (c) the specific internal energy. 5.65: Consider a throttling process where steam at 280◦ C and a quality of 96% is throttled to P2 = 200 kPa. At point 2, determine the quality (if a liquid-vapor mixture) or the temperature (if a superheated vapor). 5.66: Steam is throttled adiabatically in a steady flow from 1500 kPa and 350◦ C to 200 kPa at a flow rate of 0.01 kg/s. The inlet and outlet areas to and from the throttling device are 4 cm2 , and 24 cm2 , respectively. Determine (a) the inlet velocity, (b) the outlet velocity, and (c) the volumetric flow rate at the inlet. 5.67: Steam at 400 kPa is in a chamber equipped with a throttling calorimeter. A small quantity of steam is led to the calorimeter and is exhausted from the calorimeter (101.32 kPa) at a temperature of 150◦ C. Determine the quality of the steam in the chamber.

6 The Second Law of Thermodynamics

Chapter Objectives •

To consider the idea of quality of energy.

•

To introduce the two statements of the second law of thermodynamics (the KelvinPlanck statement and the Clausius statement) and to establish their equivalence.

To define what is meant by a reversible process and identify the causes of irreversibility. • To describe the Carnot cycle and its use as a heat engine, a refrigerator, and a heat pump. • To discuss the Carnot cycle with external irreversibilities. •

•

6.1

To develop the Kelvin and Rankine absolute temperature scales.

Introduction

In Chapter 3, heat and work were identified as two forms of energy that are of primary interest in the study of thermodynamics and it was shown that the first law of thermodynamics relates the heat and work interactions at the boundary of a system to the change of energy within the system. The first law views heat and work as quantitatively equivalent to each other but places no restrictions on the conversion of heat into work or work into heat. In other words, the first law places no constraints on the direction of conversion. In practice, however, there is a preferred direction in which processes occur. For example, a sugar cube dissolves readily in a cup of coffee but the reverse, if it happens at all, would be considered a miracle. Similarly, objects fall off a table quite readily but do not rise up against gravity on their own. Heat flows from a higher temperature to a lower temperature on its own but the reverse process requires an expenditure of work. Gases expand, liquids mix, and fuels burn quite readily but a reversal of these processes is difficult to comprehend. Indeed, conversion of work into heat (friction) is commonplace but the conversion of heat into work is not so readily accomplished. The preferential direction for the occurrences of a process is intimately linked to the idea of the quality of energy. A process occurs in a direction such that high-quality energy is degraded into low-quality energy. This is why work that is high-quality energy can readily be converted into heat that is low-quality energy but the reverse process will never be complete. This is the concept of the degradation of energy that is the essence of the second law of thermodynamics and the two classical statements of the second law of thermodynamics are the Kelvin-Planck and the Clausius statements. 149

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The Kelvin-Planck Statement and Heat Engines

The conversion of heat into work is accomplished in a device called a heat engine. Usually, heat engines are designed to operate in a thermodynamic cycle, which means that the initial and final states of the working fluid are the same. The cycle consists of a number (usually four) of processes such as constant volume, constant pressure, and isothermal processes. We discussed some of these processes of an ideal gas in Chapters 3 and 4 and another, the isentropic process will be considered in Chapter 7. For the present discussion, we represent the heat engine by the diagram shown in Figure 6.1. The engine is shown operating between two thermal reservoirs, a high-temperature reservoir or a heat source at TH and a low-temperature reservoir or a heat sink at TL . Each reservoir is assumed to have an infinite thermal capacity so that its temperature, TH or TL , is unaffected by the withdrawal or deposition of heat. In practice, these reservoirs have finite thermal capacities. Consequently, to maintain a constant reservoir temperature, the heat withdrawn from the source reservoir must be replenished by an equivalent amount of heat flow to it. For example, in a steam power plant, the heat supplied by the boiler to generate the steam is continuously replenished by the burning of the fuel. Similarly, the heat rejected into the sink must be removed if the sink temperature is to be maintained at a constant value. In a steam power plant, the heat sink may be a river that eventually receives the heat that is picked up by the cooling water as it circulates through the condenser tubes. This rejected heat is carried downstream by the river currents. The heat engine in Figure 6.1 receives Q H from a source at a high temperature, TH , and rejects heat, Q L at a low temperature, TL . The difference, Q H − Q L , is converted to work so that via the first law for the engine cycle Wnet = Q H − Q L

(6.1)

and the thermal efficiency of the engine cycle is =

Wnet Q H − QL QL = =1− QH QH QH

(6.2)

HighTemperature Reservoir, TH QH

Heat Engine Wnet

QL LowTemperature Reservoir, TL FIGURE 6.1 A heat engine taking heat, Q H , from a high-temperature reservoir at TH and rejecting heat, Q L , to a lowtemperature reservoir at TL . The work done is Wnet .

The Second Law of Thermodynamics

151 25 kJ

25 kJ 0 kJ

FIGURE 6.2 Proposed heat engine for Example 6.1a.

The Kelvin1 -Planck2 statement of the second law of thermodynamics prohibits Q L from being zero and if Q L is not equal to zero,  must be less than unity. A 100% efficient heat engine is consequently forbidden by the second law of thermodynamics and the promised classical statement that asserts this fact is The Kelvin-Planck Statement It is impossible to construct a heat engine that operates in a thermodynamic cycle and produces work while exchanging heat with a single thermal reservoir. If an attempt is made to operate a steam power plant without a condenser, thereby eliminating Q L , the work required to pump the turbine exhaust back into the boiler would exceed the turbine work output and the net work would be negative. When a condenser is employed, Q L =  0. The work required to pump the condensate to the boiler is a small fraction of the turbine work output and the net work is positive.

Example 6.1 The following data for Q H , Q L , and Wnet apply to a heat engine. For each case, determine if the heat engine is feasible in light of both the first and second laws of thermodynamics. The conditions are (a) Q H = 25 kJ, Q L = 0 kJ, and Wnet = 25 kJ (Figure 6.2), (b) Q H = 50 kJ, Q L = 20 kJ, and Wnet = 30 kJ (Figure 6.3), (c) Q H = 30 kJ, Q L = 15 kJ (Figure 6.4), and Wnet = 20 kJ, and (d) Q H = 40 kJ, Q L = 40 kJ, and Wnet = 0 kJ (Figure 6.5).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) In each case, the system is the heat engine. (a) The energy balance of the first law of thermodynamics is a modification of Equation 6.1 Q H = Q L + Wnet = 0 kJ + 25 kJ = 25 kJ The engine satisfies the first law. However, Q L = 0 kJ means that all of the heat supplied is converted into work, which is impossible according to the Kelvin-Planck statement of the second law. Hence, The engine in part (a) is not feasible ⇐ 1 William

Thompson (Lord Kelvin) (1824–1907) was a British mathematician and physicist who invented the Kelvin balance and contributed to the science of thermodynamics. 2 Max

Planck (1858–1947) was a German physicist who presented the quantum theory and introduced Planck’s constant.

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Introduction to Thermal and Fluid Engineering 50 kJ

30 kJ 20 kJ

FIGURE 6.3 Proposed heat engine for Example 6.1b.

(b) The energy balance of the first law of thermodynamics is a modification of Equation 6.1 Q H = Q L + Wnet = 20 kJ + 30 kJ = 50 kJ The engine satisfies the first law and because Q L is not zero, it does not violate the second law. However, the thermal efficiency =1−

QL 20 kJ =1− = 1 − 0.40 = 0.60 QH 50 kJ

is on the high side. As we will see in Chapter 8, thermal efficiencies of spark ignition engines lie between 25% and 30% and those of compression ignition engines are between 35% and 40%. The engine in part (b) is feasible ⇐ (c) The energy balance of the first law of thermodynamics is a modification of Equation 6.1 Q H = Q L + Wnet = 15 kJ + 20 kJ = 35 kJ =  30 kJ These numbers do not add up and the first law is violated. This makes any second law consideration a “don’t care.” The engine in part (c) is not feasible ⇐ (d) The energy balance of the first law of thermodynamics is a modification of Equation 6.1 Q H = Q L + Wnet = 40 kJ + 0 kJ = 40 kJ The engine satisfies the first law. Because, Q L =  0 kJ, the second law is also satisfied. However, the net work is zero, which implies that the device is not a heat engine. The data 30 kJ

20 kJ 15 kJ

FIGURE 6.4 Proposed heat engine for Example 6.1c.

The Second Law of Thermodynamics

153 40 kJ

0 kJ 40 kJ

FIGURE 6.5 Proposed heat engine for Example 6.1d.

indicates a heat transfer process in which 40 kJ is transferred between a hot reservoir to a cold reservoir. The system in part (d) is not a heat engine ⇐

Example 6.2 A high-temperature reservoir supplies heat, Q H1 , to heat engine 1. The heat

rejected by this engine, Q L1 , is supplied to heat engine 2 (Q L1 = Q H2 ). Heat engine 2 rejects heat, Q L2 , to a low-temperature reservoir. If the thermal efficiencies of heat engine 1 and heat engine 2 are 1 and 2 , respectively (Figure 6.6), show that the thermal efficiency, , of a heat engine receiving heat, Q H1 , and rejecting heat, Q L2 , is given by  = 1 + 2 − 1 2

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The system is the two heat engines. The thermal efficiency of heat engine 1 is given by 1 = 1 −

QH1

Q L1 Q H1

Heat Engine 1 with η1 1

W1 QL1 = QH2

2 QL2

FIGURE 6.6 Two heat engines in cascade for Example 6.2.

W2 Heat Engine 2 with η2

(a)

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Similarly, the thermal efficiency of heat engine 2 is given by 2 = 1 −

Q L2 Q H2

(b)

The thermal efficiency of the overall heat engine receiving heat, Q H1 , and rejecting heat, Q L2 , is Q L2 Q H1

=1−

(c)

Because Q L1 = Q H2 , Equation (a) may be written as 1 = 1 −

Q H2 Q H1

and this may be adjusted to Q H2 = Q H1 (1 − 1 )

(d)

With Equation (d) in Equation (b), we have 2 = 1 −

Q L2 Q H1 (1 − 1 )

(e)

and from Equation (c) Q L2 =1− Q H1

(f)

With Equation (f) in Equation (e) 2 = 1 −

1− 1 − 1

and when this is solved for , we obtain the required result  = 1 + 2 − 1 2 ⇐

6.3

The Clausius Statement: Refrigerators and Heat Pumps

It is a common observation that heat flows of its own accord from regions at higher temperature to regions at lower temperature. However, the direction of heat flow cannot be reversed without the expenditure of work. This idea is embodied in the Clausius3 statement of the second law: The Clausius Statement Without a supply of work, it is impossible to construct a refrigerator or a heat pump that operates in a thermodynamic cycle and transfers heat from a lowtemperature reservoir to a high-temperature reservoir.

3 R.

J. E. Clausius (1822–1888) was a German physicist who was a founder of thermodynamics.

The Second Law of Thermodynamics

155 HighTemperature Reservoir, TH Refrigerator or Heat Pump

QH

W

QL LowTemperature Reservoir, TL FIGURE 6.7 A refrigerator or heat pump taking heat, Q L , from a low-temperature reservoir at TL and rejecting heat, Q H , to a high-temperature reservoir at TH . The work required to drive the device is W.

A modest discussion of the refrigeration or heat pump cycle was provided in Section 3.6.3 and those remarks will now be amplified. A representation of a refrigerator or a heat pump in which work, W, is supplied to drive the flow of heat from a low-temperature reservoir at TL to a higher-temperature reservoir at TH is shown in Figure 6.7. For the refrigerator operating on an ideal vapor compression cycle, the refrigerant enters the compressor as a saturated vapor and is compressed to a higher pressure and temperature. The superheated vapor is then circulated through a condenser coil and rejects heat, Q H , to the surroundings. The refrigerant leaves the condenser as a saturated liquid and passes through a throttling valve experiencing a pressure drop and emerging as a mixture of liquid and vapor. This mixture enters the evaporator coil located inside the refrigerator, picks up heat from the refrigerated space, and is converted to saturated vapor ready to go through another cycle. Application of the first law of thermodynamics to the system in Figure 6.7 gives QL + W = Q H

or

W = Q H − QL

(6.3)

and, as we have indicated in Section 3.6.3, the performance of the refrigerator is measured in terms of the coefficient of performance (COP), which is the ratio of the heat removed from the refrigerated space to the work supplied =

QL QL 1 = = Q W Q H − QL H −1 QL

(6.4)

If the system is a heat pump, the appropriate coefficient of performance is the ratio of the heat supplied to a heated space, Q H , to the work supplied =

QH QH = = W Q H − QL

From Equations 6.4 and 6.5 we have W=

QL QH =  

1 QL 1− QH

(6.5)

156

Introduction to Thermal and Fluid Engineering Refrigerant R-134a β = 2.80 . m = 0.20 kg/s –4°C x = 0.20

Evaporator Coil

Saturated Vapor

QL FIGURE 6.8 Confguration for the evaporator coil of Example 6.3.

and solving for  gives =

QH = QL



W + QL QL



 =

   W 1 +1 = +1  QL 

or =+1

(6.6)

Example 6.3 Refrigerant R-134a enters the evaporator coil (Figure 6.8) of a refrigeration

system at a temperature of −4◦ C and a quality of 0.20 and emerges as a saturated vapor at the same temperature. The coefficient of performance of the system is 2.80. If the refrigerant flow rate is 0.20 kg/s, determine the power required to drive the system.

Solution Assumptions and Specifications (1) Steady-state and steady-flow conditions exist. (2) The system is the evaporator coil. (3) Kinetic and potential energy changes are negligible. The change in enthalpy across the evaporator is given by h = h g − (h f + xh fg ) = h g − h f − xh fg = (1 − x)h fg QH

W

House 125 W/°C

QL (a) QH W

QL (b)

FIGURE 6.9 Dual-purpose heating/cooling unit for Example 6.4.

House 125 W/°C

The Second Law of Thermodynamics

157

At −4◦ C, Table A.6 in Appendix A gives h fg = 200.15 kJ/kg. Thus, h = (1 − 0.20)(200.15 kJ/kg) = 160.12 kJ/kg ˙ L is and the rate of heat removal, Q ˙ L = mh Q ˙ = (0.20 kg/s)(160.12 kJ/kg) = 32.02 kW Equation 6.4 may be expressed in rate form as =

˙L Q ˙ W

or ˙ ˙ = Q L = 32.02 kW = 11.44 kW ⇐ W  2.80

Example 6.4 A dual-purpose heating/cooling unit (Figure 6.9) is designed to maintain

a house at 20◦ C throughout the year. The power input to the unit is 1.20 kW and the heat loss or gain is 125 W per unit temperature difference between the inside and the outside. During the winter, the unit operates as a heat pump with a COP of  = 3.80 while during the summer it serves as an air conditioner with a COP of  = 3.0. Determine (a) the maximum temperature and (b) the minimum temperature for which the unit meets the duty.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The system is the dual purpose heat/cooling unit. (a) Let Tmax denote the maximum outside temperature during the summer when the unit functions as an air conditioner (refrigerator). The rate at which the unit can extract heat from the house is, according to Equation 6.4 ˙ L = W ˙ = 3.00(1.20 kW) = 3.60 kW Q The maximum heat gain of the house is ˙ max = (125 W/◦ C)(Tmax − 20◦ C) Q ˙ L and then solved for Tmax which may be equated to Q (125 W/◦ C)(Tmax − 20◦ C) = 3600 W or Tmax = 20◦ C +

3600 W = 20◦ C + 28.8◦ C = 48.8◦ C ⇐ 125 W/◦ C

(b) Let Tmin denote the minimum outside temperature during the winter when the unit functions as a heat pump. According to Equation 6.5, the rate at which the unit can supply heat to the house is ˙ H = W ˙ = 3.80(1.20 kW) = 4.56 kW Q

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The maximum heat loss of the house is ˙ max = (125 W/◦ C)(20◦ C − Tmin ) Q ˙ H and then solved for Tmin which may be equated to Q (125 W/◦ C)(20◦ C − Tmin ) = 4560 W or Tmin = 20◦ C −

6.4

4560 W = 20◦ C − 36.5◦ C = −16.5◦ C ⇐ 125 W/◦ C

The Equivalence of the Kelvin-Planck and Clausius Statements

For the second law of thermodynamics to be absolutely general, the Kelvin-Planck and Clausius statements must be equivalent. This means that a violation of the Kelvin-Planck statement implies a violation of the Clausius statement and that a violation of the Clausius statement leads to a violation of Kelvin-Planck. For example, consider Figure 6.10 and suppose that it would be possible to transfer heat from the low-temperature reservoir at TL to the high-temperature reservoir at TH as indicated in Figure 6.10a. This is clearly in violation of the Clausius statement of the second law. If we add the engine, shown in Figure 6.10b, which is not in violation of the second law, the work done will be W = Q H − Q L . For identical values of Q H , the combination of both systems is shown in Figure 6.10c, which implies that no heat is being exchanged with the hot reservoir. Such an engine is excluded by the Kelvin-Planck statement that no engine may operate in a cycle while exchanging heat with a single reservoir. Thus, the violation of the Clausius statement has led to a violation of the Kelvin-Planck statement. On the other hand, the assumption that the heat supplied, Q H , from a high-temperature reservoir at TH can be completely converted to work, W, is a violation of the Kelvin-Planck

TH

TH

QH

QH

+ QL

TH

W = QH – QL QL

=

W = QH – QL QH – QL

TL

TL

TL

(a)

(b)

(c)

FIGURE 6.10 A violation of the Clausius statement implies a violation of the Kelvin-Planck statement. (a) A transfer of heat from TL to TH without work, which is a violation of the Clausius statement. (b) A heat engine not in violation of the Clausius or the Kelvin-Planck statement. (c) A superposition of (a) and (b) that violates the Kelvin-Planck statement.

The Second Law of Thermodynamics

159

statement. Under these circumstances, the work, W, can be converted to heat Q by means of friction. With this heat, the temperature of the body can be raised to a temperature T > TH . The net result is the transfer of heat from a body at a lower temperature, TH , to a body at higher temperature, T. This violates the Clausius statement.

6.5

Reversible and Irreversible Processes

According to the Kelvin-Planck statement of the second law, it is impossible to construct a heat engine with 100% thermal efficiency. Given the reservoir temperatures, TH and TL , the paramount question concerns the maximum efficiency that a heat engine operating between these temperatures can attain. The answer to this question is rooted in the concept of a reversible process that may be defined as: A reversible process is a process that can be reversed without any residual change in the system or its surroundings. Both the system and the surroundings must return to their original states as if the forward and reverse processes never occurred. A reversible process is an ideal process that cannot be achieved in practice. Nonetheless, the reversible process provides a yardstick for measuring the irreversibility in an actual process. The four main effects that cause a process to be irreversible are friction, heat transfer through a finite temperature difference, mixing, and combustion. •

Friction causes energy (work) to be dissipated into heat (Figure 6.11) but because heat, according to the Kelvin-Planck statement of the second law, cannot be completely converted to work, the presence of friction makes a process irreversible.

•

When heat transfer from a high temperature to a low temperature body occurs (Figure 6.12), the process cannot be reversed unless the surroundings supply some work as demanded by the Clausius statement. Because the surroundings are not restored to their original state, heat transfer through a finite temperature difference is an irreversible process.

•

When two or more gases are allowed to mix (Figure 6.13), they do so spontaneously. However, if the mixture is to be separated into its original components, work must be done. Consequently, mixing is an irreversible process.

•

The combustion of a fuel (Figure 6.14) produces a mixture of gases. The impossibility of the return of these gases to the form of the original fuel shows that combustion is an irreversible process.

Other effects that cause irreversibilities include the conversion of electrical power into heat (the I 2 R loss in conductors), unrestrained expansion of fluids, hysteresis, and inelastic deformation.

F W Ff

FIGURE 6.11 One cause of an irreversibility is friction.

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Introduction to Thermal and Fluid Engineering

T1

T2 . Q

T1 > T2

FIGURE 6.12 One cause of an irreversibility is heat transfer through a finite temperature difference.

A heat engine that operates on a thermodynamic cycle consisting only of reversible processes is a reversible heat engine and provides the maximum thermal efficiency. One such cycle is the Carnot cycle, which is considered next.

6.6

The Carnot Cycle

The Carnot cycle4 is shown, for an ideal gas in P-V coordinates, in Figure 6.15. The cycle consists of four reversible processes. •

Process 1-2: A reversible isothermal process in which the working fluid receives heat, Q H , from a high-temperature reservoir at temperature, TH .

•

Process 2-3: A reversible adiabatic expansion process during which the temperature of the working fluid decreases from TH to TL where TL is the temperature of the low-temperature reservoir.

•

Process 3-4: A reversible isothermal process in which the working fluid rejects heat, Q L , to the low-temperature reservoir at temperature, TL .

•

Process 4-1: A reversible adiabatic compression process during which the temperature of the working fluid increases from TL to TH .

The foregoing reversible processes are displayed in Figure 6.15. The thermal efficiency of the Carnot cycle may be derived by assuming, for convenience, that the working fluid is an ideal gas. •

Process 1-2: The first law energy balance of Equation 3.12 can be written as Q12 − W12 = U2 − U1 = mc v (T2 − T1 )

(6.7)

However, because the process is isothermal, T1 = T2 = TH , we find that Q12 = Q H = W12 and Equation 3.43, which gives the work done in an isothermal process, produces Q H = W12 = P1 V1 ln •

V2 V2 = mRTH ln V1 V1

(6.8)

Process 2-3: The first law energy balance of Equation 3.12 can be written as Q23 − W23 = U3 − U2 = mc v (T3 − T2 ) but because the process is adiabatic, Q23 = 0, T2 = TH and T3 = TL . Hence, W23 = mc v (TH − TL )

4 Sadi

Carnot (1796–1832) was a French physicist who formulated Carnot’s theorems in thermodynamics.

(6.9)

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161 Pull Partition

Gas-1

Gas-2

FIGURE 6.13 One cause of an irreversibility is in the mixing of gases. •

Process 3-4: The first law energy balance of Equation 3.12 can be written as Q34 − W34 = U4 − U3 = mc v (T4 − T3 ) This is another isothermal process where T3 = T4 = TL and Q34 = Q L = W34 . This time Equation 3.43 gives Q L = W34 = P3 V3 ln

•

V4 V4 = mRTL ln V3 V3

(6.10)

Process 4-1: Here, the first law energy balance of Equation 3.12 can be written as Q41 − W41 = U1 − U4 = mc v (T1 − T4 ) but because the process is adiabatic, Q41 = 0, T1 = TH and T4 = TL , and W41 = mc v (TL − TH )

(6.11)

The net work of the cycle is Wnet = W12 + W23 + W34 + W41 and use of Equations 6.8 through 6.11 shows that Wnet = mRTH ln

V2 V4 + mc v (TH − TL ) + mRTL ln + mc v (TL − TH ) V1 V3

or Wnet = mRTH ln

V2 V4 + mRTL ln V1 V3

Bang

FIGURE 6.14 One cause of an irreversibility is in combustion.

(6.12)

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Introduction to Thermal and Fluid Engineering P 1 QH

TH 2

4 TL

QL 3 V

FIGURE 6.15 Four reversible processes in a Carnot cycle.

The heat into the cycle is given by Equation 6.8 Q H = mRTH ln

V2 V1

(6.8)

The thermal efficiency of the Carnot cycle, C , will be Wnet C = = QH

V2 V4 + mRTL ln V1 V3 V2 mRTH ln V1

mRTH ln

or V4 V3 V2 ln V1 ln

TL C = 1 + TH

(6.13)

It will be shown in the next chapter that for an ideal gas executing a reversible adiabatic process, T V k−1 = C

(a constant)

(6.14)

and when we apply this to processes 2-3 and 4-1, which are reversible adiabatic processes, we obtain  k−1 V2 TL = (6.15) V3 TH and 

V1 V4

k−1 =

TL TH

(6.16)

From these, it follows that V2 V3 = = V1 V4



V4 V3

−1 (6.17)

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163

P 1 TH

QH 2 4 TL

QL

3 V

FIGURE 6.16 The P-V diagram for the reversed Carnot cycle showing four reversible processes.

Then, after substitution of Equation 6.17 into Equation 6.13, the efficiency of the Carnot cycle is seen to be TL TH

C = 1 −

(6.18)

Even though this efficiency has been derived for an ideal gas, it applies to every reversible engine operating between the temperature limits of TH and TL . When we compare Equations 6.2 and 6.18, we find that for a Carnot cycle and indeed for any reversible cycle that QL TL = QH TH

(6.19)

Equation 6.19 will be employed in the next chapter when we identify a new property known as entropy. The reversed Carnot cycle (shown in Figure 6.16 with the directions of Q L and Q H reversed) can either represent a refrigeration or a heat pump cycle. The performance of a refrigerator is measured in terms of the coefficient of performance and for the refrigerator, the coefficient of performance was defined by Equations 3.19 and 6.4. For the heat pump, the coefficient of performance was defined by Equations 3.20 and 6.5. With Q L /Q H replaced by TL /TH in accordance with Equation 6.19 in Equations 6.4 and 6.5, we have for the Carnot refrigerator C =

1 TH −1 TL

(6.20)

and for the Carnot heat pump C =

1 1−

TL TH

(6.21)

Example 6.5 A Carnot heat pump (Figure 6.17) supplies heat to a steam generator where water enters as saturated liquid at 100 kPa and exits as saturated vapor at the same pressure. The heat pump consumes 80 kW of power and draws heat from the environment at 10◦ C. Determine the rate of steam generation.

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Introduction to Thermal and Fluid Engineering Saturated Liquid

Steam Generator

100 kPa

Saturated Vapor 100 kPa

Heat Pump 80 kW

10°C (283 K) FIGURE 6.17 Configuration for Example 6.5.

Solution Assumptions and Specifications (1) All processes in the cycle are internally reversible. (2) The control volume is the steam generator. (3) The system is the steam in the steam generator. (4) There are no potential and kinetic energy changes. (5) The system operates in the steady state. From Table A.4 at 100 kPa, read Tsat = 99.6◦ C

and

h fg = 2258.0 kJ/kg

Then with TH = 99.6◦ C + 273◦ C = 372.6 K

and

TL = 10◦ C + 273◦ C = 283 K

Equation 6.21 gives C =

1 TL 1− TH

=

1 1 = = 4.16 283 K 1 − 0.760 1− 372.6 K

˙ = 80 kW For W ˙ H = C W ˙ = (4.16)(80 kW) = 332.7 kW Q and m ˙ =

˙H Q 332.7 kW = = 0.147 kg/s ⇐ h fg 2258.0 kJ/kg

Example 6.6 A three-process power cycle employing an ideal gas with constant specific heats is shown in Figure 6.18. It consists of the following: •

Process 1-2: A constant volume heat addition

•

Process 2-3: An adiabatic expansion with k = c p /c v = 1.40

•

Process 3-1: A constant pressure heat rejection with V3 /V1 = 2

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165

P 2

PV k =

QH

C

1

3 QL

V FIGURE 6.18 Three processes for Example 6.6.

(a) Show that the thermal efficiency of the cycle is given by ⎡ ⎤ V3 − 1 ⎥ Wnet T1 ⎢ ⎢ V1 ⎥ C = =1−k T1 ⎦ QH T2 ⎣ 1− T2 and then compute, with TH = 900 K and TL = 300 K, (b) the cycle efficiency and (c) the Carnot efficiency.

Solution Assumptions and Specifications (1) All processes in the cycle are internally reversible. (2) An ideal gas with constant specific heats is specified. (3) Steady flow and steady state applies. (a) Application of the first law stated by Equation 3.12 to process 1-2 gives Q12 − W12 = U2 − U1 = mc v (T2 − T1 ) and because for this constant volume process, W12 = 0 Q12 = Q H = mc v (T2 − T1 ) For process 2-3, the first law gives Q23 − W23 = U3 − U2 = mc v (T3 − T2 ) and because for this adiabatic process, Q23 = 0 W23 = mc v (T2 − T3 ) For process 3-1, the first law provides Q31 − W31 = U1 − U3 = mc v (T1 − T3 ) and for a constant pressure process where W31 = P1 V1 − P3 V3 , we may write Q31 = mc p (T1 − T3 ) + P1 V1 − P3 V3 = H1 − H3

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Introduction to Thermal and Fluid Engineering

or Q L = −Q31 = mc p (T3 − T1 ) The thermal efficiency of the cycle may now be expressed as   T3 T1 −1 mc p (T3 − T1 ) QL T1  =1− =1− =1−k  T1 QH mc v (T2 − T1 ) T2 1 − T2 Because the ideal gas law requires that, at constant pressure, T3 /T1 = V3 /V1 , we have ⎛ ⎞ V3 − 1 ⎟ T1 ⎜ ⎜ V1 ⎟ ⇐ C = 1 − k ⎝ T1 ⎠ T2 1− T2 and this is the required result. (b) We note that T1 = TL = 300 K and T2 = TH = 900 K. Then with V3 /V1 = 2.0, ⎛ ⎞ ⎛ ⎞ V3   − 1 ⎟ T1 ⎜ 2−1 ⎟ ⎜ V1 ⎟ = 1 − (1.40) 300 K ⎜ C = 1 − k ⎝ ⎠ = 0.30 ⇐ ⎝ ⎠ T 300 K T2 900 K 1 1− 1− T2 900 K (c) For a Carnot engine C = 1 −

6.7

TL 300 K =1− = 0.667 TH 900 K

(66.7%) ⇐

The Carnot Cycle with External Irreversibilities

Isothermal heat addition and heat rejection processes are not achievable in practice. A more feasible approach is to assume that the heat is supplied from a reservoir at a source temperature, Tsource , through a heat exchanger of surface area, SH , with an overall heat transfer coefficient, UH , such that ˙ H = UH SH (Tsource − TH ) Q

(6.22)

Because the heat supply involves a finite temperature difference, Tsource − TH , the pro˙ L , is assumed to be isothermal at a cess is externally irreversible. If the heat rejection, Q temperature, TL , then the power output, Wnet will be   ˙ net = C Q ˙ H = 1 − TL UH SH (Tsource − TH ) W (6.23) TH ˙ net can be optimized if and W ˙ net dW =0 dTH

(6.24)

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167

The result of the indicated differentiation, upon simplification, will be TH = (Tsource TL ) 1/2

(6.25)

and the efficiency of the cycle then becomes TL TL C = 1 − =1− =1− TH (Tsource TL ) 1/2



1/2

TL Tsource

(6.26)

On the other hand, if the heat addition is isothermal but the heat rejection is to a heat sink at temperature, Tsink , through a heat exchanger of surface area, SL , via a heat transfer coefficient of UL , then ˙ L = UL SL (TL − Tsink ) Q

(6.27)

   TH TH TL ˙ ˙ ˙ Wnet = C Q H = C Q L = 1− UH SH (TL − Tsink ) TL TH TL

(6.28)

and

˙ net and TL will require setting d W ˙ net /dTL = 0 Now the optimization procedure involving W and the result of the indicated differentiation, upon simplification, will be TL = (Tsink TH ) 1/2

(6.29)

and the efficiency of the cycle becomes TL (Tsink TH ) 1/2 C = 1 − =1− =1− TH TH



Tsink TH

1/2 (6.30)

When both the heat addition and the heat rejection involve heat transfers through finite temperature differences, Tsource − TH and TL − Tsink respectively, then it can be shown that ˙ net is at its optimum value when W TL = TH



Tsink Tsource

1/2 (6.31)

in which case, the cycle efficiency will be  C = 1 −

Tsink Tsource

1/2 (6.32)

Example 6.7 A Carnot engine operates between a heat source at 1200 K and a heat sink at 300 K. The heat supply rate to the engine is 100 kW. Determine the temperatures TH and TL , the engine efficiency, the net power, and the heat rejection rate for (a) a heat supply through a finite temperature difference but an isothermal heat rejection, (b) an isothermal heat supply but heat rejection through a finite temperature difference, (c) both the heat supply and the heat rejection occurring through a finite temperature difference, and (d) the engine operates on a Carnot cycle with no external irreversibilities.

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Introduction to Thermal and Fluid Engineering

Solution Assumptions and Specifications (1) Wnet is optimized in each case. (2) TH = 1200 K and TL = 300 K are specified. (3) The control volume is a Carnot engine. (a) For a heat supply through a finite temperature difference but an isothermal heat rejection, the temperature, TL equals the sink temperature Tsink = TL = 300 K ⇐ The temperature, TH is given by Equation 6.25 with Tsource = 1200 K TH = (Tsource TL ) 1/2 = [(1200 K)(300 K)]1/2 = 600 K ⇐ Equation 6.26 gives the thermal efficiency  C = 1 −

1/2

TL

 =1−

Tsource

300 K 1200 K

1/2 = 0.50

(50%) ⇐

The net engine power will be ˙ net = C Q ˙ H = (0.50)(100 kW) = 50 kW ⇐ W and the heat rejected derives from the first law ˙L=Q ˙H−W ˙ net = 100 kW − 50 kW = 50 kW ⇐ Q (b) For an isothermal heat supply but heat rejection through a finite temperature difference, the temperature, TH equals the source temperature Tsource = TH = 1200 K ⇐ The temperature, TL is given by Equation 6.29 with Tsink = 300 K TL = (Tsink TH ) 1/2 = [(300 K)(1200 K)]1/2 = 600 K ⇐ Equation 6.30 gives the thermal efficiency  C = 1 −

Tsink TH

1/2

 =1−

300 K 1200 K

1/2 = 0.50

(50%) ⇐

The net engine power will be ˙ net = C Q ˙ H = (0.50)(100 kW) = 50 kW ⇐ W and the heat rejected derives from the first law ˙L=Q ˙H−W ˙ net = 100 kW − 50 kW = 50 kW ⇐ Q (c) For both the heat supply and the heat rejection occurring through a finite temperature difference, we have Tsource = 1200 K and Tsink = 300 K. We cannot find TH and TL unless UH SH or UC SC are known. However, their ratio may be obtained from Equation 6.31     TL Tsink 1/2 300 K 1/2 1 = = = TH Tsource 1200 K 2

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169

The thermal efficiency may be found from Equation 6.32     Tsink 1/2 300 K 1/2 C = 1 − =1− = 0.50 Tsource 1200 K

(50%) ⇐

The net engine power will be ˙ net = C Q ˙ H = (0.50)(100 kW) = 50 kW ⇐ W and the heat rejected derives from the first law ˙L=Q ˙H−W ˙ net = 100 kW − 50 kW = 50 kW ⇐ Q (d) For a Carnot cycle with no external irreversibilities, TH = Tsource = 1200 K and TL = Tsink = 300 K. The thermal efficiency is given by C = 1 −

TL 300 K =1− = 0.75 TH 1200 K

(75%)

The net engine power will be ˙ net = C Q ˙ H = (0.75)(100 kW) = 75 kW ⇐ W and the heat rejected derives from the first law ˙L=Q ˙H−W ˙ net = 100 kW − 75 kW = 25 kW ⇐ Q

6.8

The Absolute Temperature Scales

Equation 6.19 that applies to every reversible heat engine provides a basis for the absolute Kelvin temperature scale. Because Equation 6.19 defines only a ratio of temperatures, a reference state must be assigned. The reference state selected is the triple point of water that is assigned a value of 273.16 K. If a reversible heat engine receives heat, Q, from a reservoir at the reference temperature and rejects heat, Qref , to a reservoir at the reference temperature, Tref , then T = Tref

Q Q = 273.16 Qref Qref

(6.33)

Equation 6.33 defines the absolute Kelvin temperature scale. The scale is independent of the working fluid used in a reversible engine. The Rankine temperature scale is defined in terms of the Kelvin temperature scale. The relationship between the two scales is linear and is given by T( ◦ R) = 1.8T( K )

6.9

(6.34)

Summary

The preferential direction for the occurrence of a thermodynamic process is intimately linked to the idea of the quality of energy and this is the essence of the second law of thermodynamics.

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Introduction to Thermal and Fluid Engineering

Two classical statements of the second law of thermodynamics are the Kelvin-Planck statement It is impossible to construct a heat engine that operates in a thermodynamic cycle and produces work while exchanging heat with a single thermal reservoir. and the Clausius statement Without a supply of work, it is impossible to construct a refrigerator or a heat pump that operates in a thermodynamic cycle and transfers heat from a lowtemperature reservoir to a high-temperature reservoir. These statements are equivalent and a violation of one of them infers a violation of the other. The performance of both refrigeration and heat pump cycles is measured by the coefficient of performance, abbreviated COP. For the refrigerator, the coefficient of performance is 1 QH −1 QL

=

and for the heat pump, the coefficient of performance is =

1 QL 1− QH

The relationship between  and  is =+1 A reversible process is a process that can be reversed without any residual change in the system or its surroundings. Examples of irreversible processes include the following: •

Motion with friction

•

Unrestrained expansion

•

Mixing and diffusion

•

Spontaneous chemical reactions

•

Heat transfer through a finite temperature difference

•

Current flow through an electrical resistor

The Carnot cycle is composed of four reversible processes working between the temperatures, TH , and, TL . The efficiency of a Carnot engine is C = 1 −

TL TH

The coefficient of performance of the Carnot refrigerator is C =

1 TH −1 TL

The Second Law of Thermodynamics

171

and the coefficient of performance of the Carnot heat pump is C =

6.10

1 1−

TL TH

Problems

Heat Engines 6.1: A steam power plant generates 5 MW of electrical power while burning natural gas at a rate of 420 kg/s in the boiler. The plant rejects heat to a condenser in which the cooling water enters at 27◦ C and leaves at 35◦ C. Assume the heating value of the natural gas to be 30,000 kJ/kg and the specific heat of water to be c = 4.184 kJ/kg-K. Determine (a) the thermal efficiency and (b) the mass flow rate of the cooling water required. 6.2: The following data for Q H , Q L , and Wnet apply to a heat engine. For the heat engine to be feasible, both the first and second laws of thermodynamics must be satisfied. For each case, determine whether the heat engine is feasible. The cases are (a) Q H = 200 kJ, Q L = 90 kJ, and Wnet = 110 kJ, (b) Q H = 50 kJ, Q L = 0, and Wnet = 35 kJ, and (c) Q H = 18 kJ, Q L = 18 kJ, and Wnet = 0. 6.3: Two heat engines operate in series. Engine 1 receives heat at the rate of 20 kW and produces a power of 8 kW. The heat rejected by engine 1 is used to drive engine 2, which produces 4 kW. Determine (a) the thermal efficiency of engine 1, (b) the thermal efficiency of engine 2, and (c) the thermal efficiency of the combined system. 6.4: Figure P6.4 shows two heat engines, one operating between reservoirs at absolute temperatures T1 and T2 with efficiency 1 and the other operating between reservoirs at absolute temperatures T2 and T3 with efficiency 2 . Show that (a) Q3 = (1 − 1 )(1 − 2 ) Q1 T1 Heat Engine 1 W1

T2 Heat Engine 2 W2

T3 FIGURE P6.4

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Introduction to Thermal and Fluid Engineering and (b) W1 + W2 = (1 + 2 − 1 2 ) Q1

6.5: A gasoline engine delivers a power output of 20 kW and has a thermal efficiency of 28% based on a the heating value of gasoline of 44,000 kJ/kg. Determine (a) the rate of fuel consumption in the engine and (b) the power lost through the exhaust and the radiator. 6.6: An engine uses fuel oil at the rate of 0.1 kg/s and produces 850 W. The heating value of the oil is 41,850 kJ/kg. Determine (a) the heat rejection from the engine and (b) its thermal efficiency. 6.7: A power cycle uses an ideal gas with constant specific heat as the working medium. The cycle consists of four processes: Process 1-2: an isothermal compression Process 2-3: a constant volume heat addition Process 3-4: an isothermal expansion Process 4-1: a constant volume heat rejection Draw a P-V diagram and develop an expression for the thermal efficiency of the cycle in terms of R, c v , T1 , T3 , V1 , and V3 . 6.8: Two reversible engines are connected between a heat source and a heat sink, as indicated in Figure P6.4. In addition, 1 = 2 , T1 = 580 K and T3 = 220 K and the heat entering engine 1 is 120 W. Determine (a) T2 , (b) W1 , (c) W2 , and (d) the heat rejected by engine 2. 6.9: The thermal efficiency of a particular engine operating on an ideal cycle is 35%. Determine (a) the heat supplied per 1.5 kW of work developed, (b) the ratio of the heat supplied to the heat rejected, and (c) the ratio of the work developed to the heat rejected. 6.10: A thermodynamic cycle has a steam generator to which 72 kW are added. The working fluid drives an engine that produces 27 kW. The pump that circulates the working medium absorbs 2 kW. Determine (a) the net work, (b) the heat rejected, and (c) the thermal efficiency. 6.11: A heat engine draws in heat at the rate of 100 kW from a high-temperature source. If the thermal efficiency of the heat engine is 28%, determine the rate at which heat is rejected to the low-temperature sink. 6.12: A coal burning power plant produces 30 MW of power. Each kilogram of coal yields 6000 kJ of energy when burned, and the plant has an efficiency of 30%. Determine the coal consumption of the plant in kg/h. 6.13: Determine the unknown quantity in each case in the following table: Case QH , kJ (1) (2) (3) (4)

100 250 (e) 1000

QL , kJ ,% Wnet , kJ (a) 200 (f) (g)

(b) (c) 28 32

35 (d) 60 (h)

6.14: The following data for Q H , Q L , and  apply to a heat engine. For each case, determine whether the heat engine is feasible and, if feasible, determine the heat output.

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173

˙ H , kW Q ˙ L , kW ,% Case Q (1) (2) (3) (4)

300 500 800 200

150 300 400 150

60 40 65 25

6.15: Three heat engines operate in series. Engine 1 receives heat at the rate of 100 kW and produces 25 kW of power. The heat rejected by engine 1 is received by engine 2, which produces a power of 20 kW. The heat rejected by engine 2 is received by engine 3, which produces a power of 15 kW. Determine (a) the thermal efficiency of each engine, (b) the heat rejected by each engine, and (c) the thermal efficiency of the combined system. 6.16: For the three-engine arrangement shown in Figure P6.16, show that the thermal efficiency, , is  = (1 − 1 )(1 − 2 )(1 − 3 ) where 1 , 2 , and 3 are the respective thermal efficiencies of the individual engines

Q1 1

W1 Q2

2

W2 Q3

3

W3 Q4

FIGURE P6.16

6.17: A steam power plant having a thermal efficiency of 35% generates 25 MW of power. The plant discharges heat in a condenser that draws cooling water at the rate of 820 kg/s from a river at 25◦ C. Determine the temperature of the water after it circulates through the condenser. 6.18: Two heat engines operate in parallel between the same source and sink as shown in Figure P6.18. Determine (a) the work outputs, W1 and W2 , (b) the thermal efficiencies, 1 and 2 , and (c) the thermal efficiency of the combined system.

174

Introduction to Thermal and Fluid Engineering Source 200 kJ

100 kJ 2

1

W1 90 kJ

W2 70 kJ

Sink FIGURE P6.18

6.19: Consider the arrangement shown in Figure P6.19 and prove that the thermal efficiency of the combined arrangement, , is given by 1 + 2 = 2 where 1 and 2 are the respective thermal efficiencies of engines 1 and 2 and Q H1 = Q H2 = Q H . Source QH2

QH1 W1

2

1 QL1

W2 QL2

Sink FIGURE P6.19

6.20: A diesel engine with a thermal efficiency of 26% provides a power output of 220 kW. If the engine consumes 80 kg/h of diesel fuel, determine (a) the heating value of the fuel in kJ/kg and (b) the power lost to the environment. Refrigerators and Heat Pumps 6.21: In a vapor compression refrigeration system using refrigerant R-134a, heat is extracted from the refrigerated space as the refrigerant circulates through the evaporator and is rejected to the surroundings when the refrigerant flows through the condenser. In one such system, the R-134a enters the evaporator at −8◦ C with a quality of 0.30 and leaves the evaporator as saturated vapor at −8◦ C. The refrigerant enters the condenser at 0.8 MPa and 40◦ C and leaves the condenser as saturated liquid at 0.8 MPa. Determine (a) the coefficient of performance of the system and (b) the work input to the system. 6.22: A 1-ton window air conditioner draws energy from an electric outlet at the rate of 1.20 kW. The cost of electricity is 0.25/kW-h and 1 ton of refrigeration is equivalent to 211 kJ/min. Determine (a) the coefficient of performance of the air conditioner, (b) the heat rejected to the surroundings, and (c) the cost to operate the unit over 30 days with operation for 6 h/day.

The Second Law of Thermodynamics

175

6.23: A refrigerant enters the evaporator coil of a refrigeration system with a quality of 0.20 and leaves as saturated vapor without experiencing any pressure loss. The specific enthalpies of saturated liquid and saturated vapor are 157.31 kJ/kg and 1436.7 kJ/kg, respectively. The system incorporates a compressor that requires 300 kJ/kg of work. Determine the coefficient of performance of the system. 6.24: A household refrigerator with a coefficient of performance of 2.60 is able to cool its contents from a temperature of 20 to 10◦ C in 2 h. The mass of the contents may be assumed to be 10 kg and their average specific heat is taken as 3.10 kJ/kg-K. Determine (a) the rate at which heat is rejected by the refrigerator to the surrounding air and (b) the power input to the compressor of the refrigerator. 6.25: Figure P6.25 shows two refrigerators, one operating between reservoirs at absolute temperatures T1 and T2 with a coefficient of performance of 1 and the other operating between reservoirs at absolute temperatures T2 and T3 with a coefficient of performance of 2 . Show that the coefficient of performance between reservoirs at absolute temperatures T1 and T3 is given by =

1 2 1 + 1 + 2

T1

β1

W1

T2

β2

W2

T3 FIGURE P6.25

6.26: A heat pump draws heat from the outside air and delivers it to the inside of a house that is maintained at 20◦ C. The heat loss from the house is estimated to be 40,000 kJ/h. The heat pump has a coefficient of performance of 3.50. Determine (a) the power required to drive the heat pump, (b) the rate at which heat is extracted from the outside air, (c) the cost of operating the heat pump for 8 h if the cost of electricity is 20 cents per kW-h, and (d) the cost of heating the house for 8 h if resistance heating is used instead of the heat pump. 6.27: Figure P6.27 shows two heat pumps, one operating between reservoirs at absolute temperatures T1 and T2 with a coefficient of performance of 1 and the other operating between reservoirs at absolute temperatures T2 and T3 with a coefficient of performance of 2 . Show that the coefficient of performance between reservoirs at absolute temperatures T1 and T3 is given by =

1 2 1 + 2 − 1

176

Introduction to Thermal and Fluid Engineering T1

γ1

W1

T2

γ2

W2

T3 FIGURE P6.27

6.28: For a cooling load of 2.6 kW, determine (a) the power required to drive a refrigeration unit with a coefficient of performance of 3.50 and (b) the coefficient of performance if the unit is used as a heat pump. 6.29: A refrigeration unit is required to handle a cooling load of 15,000 kJ/h. The unit is driven by a 1.2-kW electric motor. Determine its coefficient of performance. 6.30: Refrigerant R-134a enters the evaporator of a refrigerator at −8◦ C with a quality of x = 0.22 and leaves the evaporator as saturated vapor at −8◦ C. The volumetric flow rate of the refrigerant is 4.23 × 10−3 m3 /s at the entrance of the evaporator. The refrigerator consumes 12 kW of power. Determine its coefficient of performance. 6.31: The refrigeration unit of a food storage room consumes 3.2 kW and operates with a coefficient of performance of 2.96. Determine (a) the rate of heat removal from the room and (b) the heat rejected to the environment in which the unit is located. 6.32: An air-conditioning unit consumes 10 kW of power and removes 860 kJ/m of heat from the air-conditioned space. Determine (a) the coefficient of performance of the unit, (b) the heat rejected to the environment in kilojoules over a period of 2 h, and (c) the daily cost of electricity to run the unit if the unit operates for 8 h and the cost of electricity is $0.20 per kW-h. 6.33: The contents of a household refrigerator have a mass of 12 kg with an average specific heat of 3.20 kJ/kg-K. When the refrigerator is turned on, the contents are initially at 25◦ C. The refrigerator has a coefficient of performance of 2.75 and draws 600 W of power. Determine how long it will take to cool the contents from 25 to 12◦ C. 6.34: An air-conditioning system with a coefficient of performance of 3.20 is used to maintain the temperature of a dwelling at 22◦ C. Heat leakage into the dwelling from the surroundings is at the rate of 5 kW. Heat generated owing to the occupants and several electrical appliances located in the dwelling amounts to 2 kW. Determine the power consumption of the system. 6.35: A dual-purpose heating/cooling unit is designed to maintain a building at 20◦ C throughout the year. The unit consumes 20,000 kJ/h and the heat loss/gain to or from the building is 1800 kJ/h per unit of temperature difference between the inside and outside of the building. During the summer months, the unit operates as an air conditioner with a coefficient of performance of 3.20; during the winter months, it

The Second Law of Thermodynamics

177

operates as a heat pump with a coefficient of performance of 4.00. Determine the maximum and minimum outside temperatures for satisfactory operation of the unit. 6.36: For the two-refrigerator arrangement shown in Figure P6.36, determine a relationship between COP1 , COP2 , and the combined COP for the entire system. Here COP1 is the coefficient of performance of refrigerator 1 and COP2 is the coefficient of performance of refrigerator 2. High-Temperature Reservoir Q3

COP2 2

W2 Q2

COP1

W1

1 Q1

Low-Temperature Reservoir

FIGURE P6.36

6.37: Determine the unknown quantity for each refrigeration unit in the following table:

Case

˙ H , kW Q

˙ L , kW Q

COP

˙ kW W

(1) (2) (3) (4)

120 220 (e) (g)

(a) 160 200 (h)

3.10 (c) (f) 3.20

(b) (d) 80 60

6.38: Determine the unknown quantity for each of the cases in Problem 6.37 if the data are for a heat pump. 6.39: Reconsider Problem 6.36 and assume that the arrangement pertains to two heat pumps. Determine a relationship between COP1 , COP2 , and the combined COP for the entire system. Here COP1 is the coefficient of performance of refrigerator 1 and COP2 is the coefficient of performance of refrigerator 2. 6.40: A heat pump with a coefficient of performance of 3.8 is used to heat a house in winter and maintain a temperature of 20◦ C for comfort. The heat loss from the house to the outside environment occurs at the rate of 23,600 kJ/h. Determine the power required (in kW) to drive the heat pump. 6.41: For the two-heat pump arrangement shown in Figure P6.41, show that COP1-2 =

2COP1 COP2 COP1 COP2

178

Introduction to Thermal and Fluid Engineering High-Temperature Reservoir Q2 2

1

W1

Q2

Q1

W2 Q1

Low-Temperature Reservoir FIGURE P6.41

The Carnot Cycle 6.42: A Carnot engine receives heat at a rate of 50 kW and rejects waste heat to a sink at 300 K. The efficiency of the engine is 0.55. Determine (a) the temperature of the source and (b) the power to the engine. 6.43: The hot reservoir of a Carnot engine is a combustion chamber that burns fuel at the rate of 2 kg/s with products of combustion at 1200 K. The exhaust from the engine is at 400 K and is rejected to the atmosphere. If the heating value of the fuel is 30,000 kJ/kg, find (a) the maximum power output of the system and (b) the maximum thermal efficiency attainable. 6.44: Figure P.6.44 shows a reversible heat pump drawing heat from a reservoir at absolute temperature, T2 and delivering it to a reservoir at absolute temperature, T1 . The heat pump is driven by a reversible heat engine that operates between two reservoirs at absolute temperatures T3 and T4 . Show that the ratio of heat delivered by the heat pump, Q1 , to the heat drawn by the heat engine, Q3 , is given by Q1 1 − (T4 /T3 ) = Q3 1 − (T2 /T1 )

T1

T3 Q1

Heat Pump

Q3 Heat Engine

W Q2 T2

Q4 T4

FIGURE P6.44

6.45: Consider the two heat pumps shown in Figure P6.45 with T1 = 293 K, T2 = 263 K, and ˙ 1 = 10 kW. The heat pumps are reversible with identical coefficients of performance. Q Determine (a) the coefficient of performance of each heat pump, (b) the temperature, ˙ 2 , and (d) Q ˙ 3. T2 , of the intermediate reservoir, (c) Q

The Second Law of Thermodynamics

179 T1 = 293 K . Q1 = 10 kW . W1

1 . Q2

r1 = r2

T2 . Q2

. W2

2 . Q3 T3 = 263 K FIGURE P6.45

6.46: Consider the two heat engines shown in Figure P6.46. Assume the heat engines to be reversible and to have identical efficiencies with T1 = 800 K and T3 = 300 K. Determine (a) the thermal efficiency of each engine, (b) the thermal efficiency of the two engines working together, and (c) the work output of the two engines if Q1 = 500 kJ. T1 = 800 K Q1= 500 kJ W1 T2 W2

T3 = 300 K FIGURE P6.46

6.47: A Carnot engine draws heat Q1 from a source at absolute temperature T1 and rejects heat to two sinks, Q2 to a sink at absolute temperature T2 and Q3 to a sink at absolute temperature T3 . Show that for Q2 = Q3 /2, the thermal efficiency of the engine is given by =1−

3T2 T3 T1 (2T2 + T3 )

6.48: A Carnot heat pump draws heat from two reservoirs, Q2 from a source at absolute temperature, T2 , and Q3 from a source at absolute temperature, T3 . The heat pump delivers heat, Q1 , to a reservoir at absolute temperature, T1 . Show that for Q2 = Q3 /3, the coefficient of performance of the heat pump is given by =1−

1 4T2 T3 1− T1 (3T2 + T3 )

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Introduction to Thermal and Fluid Engineering

6.49: A Carnot engine draws one-third of its heat supply from a source at 1000 K and twothirds from a source at 800 K. The engine rejects half of its heat to a sink at 400 K and the other half to a sink at 300 K. Determine the thermal efficiency of the engine. 6.50: A Carnot refrigerator extracts heat from the interior of a refrigerated space at −10◦ C and discharges it to a room at 20◦ C. The working fluid is refrigerant R-134a, which circulates through the system at the rate of 0.40 kg/s. The refrigerant experiences a change in enthalpy of 180 kJ/kg as it flows through the evaporator coil. Determine (a) the coefficient of performance of the refrigerator, (b) the power input to the refrigerator, (c) the rate of heat rejection to the surroundings, and (d) the pressure and temperature of the refrigerant entering the condenser. 6.51: The following data pertain to four heat engines. Determine whether each engine is reversible, irreversible, or impossible: Case (1) (2) (3) (4)

QH , kJ

QL , kJ

TH K

TC K

η

300 600 400

1000 1200 2600 800

600 700 1200 300

0.72

500 700 800

6.52: A reversible heat engine, with a thermal efficiency of 70%, rejects heat to a sink at 400 K. Determine (a) the temperature of the source and (b) the work output of the engine. 6.53: A heat pump uses R-134a as the working fluid and provides heating at a rate of 16 kW. Saturated vapor at 2 bar leaves the evaporator and a liquid-vapor mixture leaves with a quality of x = 0.20 leaves the condenser at 8 bar. Determine (a) the coefficient of performance and (b) the power required to drive the system. 6.54: A reversible heat engine operates between reservoirs at temperatures, TH and TS as indicated in Figure P6.54. The engine drives a reversible refrigerator that operates between temperatures, TC and TS . Show that QC 1 = QH Ts



1 1 − TC TS

  TS 1− TH

TS TH Engine

Refrigerator

W

TC Both the Engine and the Refrigerator Are Reversible FIGURE P6.54

The Second Law of Thermodynamics

181

6.55: A refrigerator extracts heat, Q L , from a reservoir at 268 K and rejects heat, Q H , to a reservoir at 300 K. For each of the following, determine whether the refrigerator operates reversibly, irreversibly, or is impossible: (1) (2) (3) (4)

Q L = 670 kJ and W = 80 kJ. Coefficient of performance (COP) = 9. Q H = 1200 kJ and Q L = 800 kJ. Q H = 1000 kJ and W = 200 kJ.

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7 Entropy

Chapter Objectives •

To introduce entropy from both statistical and classical thermodynamic perspectives.

•

To derive the Clausius inequality.

To develop and illustrate the use of the temperature-entropy diagram. • To consider the Gibbs or TdS relations for a simple compressible substance. •

•

To present equations for the entropy change of solids, liquids, and gases.

•

To establish the isentropic relations for an ideal gas.

•

To define the isentropic efficiencies for turbines, compressors, pumps, and nozzles.

•

To develop entropy balance equations for closed and open systems.

7.1

Introduction

Chapter 6 emphasized that the preferred direction for the occurrence of a process is one in which high-quality energy degrades into low-quality energy. This principle is akin to the idea of molecular disorder. The idea of disorder is a familiar one as things always seem to become disordered spontaneously but never return to an orderly arrangement on their own. In 1868, Rudolf Clausius1 suggested that the property entropy be used as a measure of molecular disorder within a system and Ludwig Boltzmann2 related the concept of entropy to the probability or number of ways in which the state of a system can be formulated. Thus, a substance has a lower value of entropy when it is in the solid phase because the number of ways in which the molecules can arrange themselves in a solid is limited. On the other hand, a substance in a gaseous phase has a higher entropy value because there are a very large number of ways in which the molecules can arrange themselves. So we are led to the fact that, as the number of ways that a state may arise increases, the more probable the state and the more disordered the system in that state.

1 Rudolf

Julius Emmanuel Clausius (1822–1888) was a German physicist and a founder of thermodynamics.

2 L.

E. Boltzmann (1844–1906) was an Austrian physicist who was a pioneer in the development of the kinetic theory of gases.

183

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Introduction to Thermal and Fluid Engineering

Based on the foregoing rationale, the entropy, S, of a system has been defined in statistical thermodynamics via the Boltzmann equation S = k ln P

(7.1)

where P is the thermodynamic probability (or the total number of quantum states available to the system) and k is the Boltzmann constant. Although in this text, our approach to thermodynamics is classical, we have briefly introduced the statistical approach in order to provide the reader with a more fundamental insight.

7.2

The Classical Definition of Entropy

The thermodynamic analysis of Carnot cycles, presented in Section 6.6, led to Equation 6.19 that may be rearranged to yield QH QL − =0 TH TL

(7.2)

We note that Q H and Q L are reversible heat transfers and Equation 7.2 reveals that, in a Carnot cycle, the algebraic sum of the reversible heat transfers divided by their corresponding absolute temperatures is zero. Because the initial and final states in a cycle are the same, the net (algebraic) change in every property in the system undergoing the cycle is zero. It follows that the quantity, Qrev /T, whose algebraic sum is zero, as reflected by Equation 7.2, must be a property of the system. This is the property entropy, S, whose definition from a statistical perspective was given by Equation 7.1. From a classical point of view, the differential change in entropy, d S, is defined as   Q dS = (7.3) T rev and the change in entropy, S2 − S1 , may be obtained by integrating Equation 7.3 between points 1 and 2 giving   2 Q S2 − S1 = (7.4) T rev 1 We observe that S is an extensive property whose units are kJ/K. The changes in specific entropy, s, with units kJ/kg-K are  q  ds = (7.5) T rev and



2

s2 − s1 = 1

 q  T

rev

(7.6)

where q is the specific reversible heat transfer in kJ/kg. Each of Equations 7.3 through 7.6 shows that, for a reversible adiabatic process, (i.e., for q = 0), the change in entropy is zero. Such a process is called an isentropic process and it may be recalled that two such processes (processes 2-3 and 4-1 in Figure 6.15) were used in constructing the Carnot cycle. It is important to recognize that entropy is a property and the change in entropy between state 1 and state 2 is independent of the path between the two states. However,

Entropy

185

the calculation of S2 − S1 using Equation 7.4 or s2 − s1 , using Equation 7.6, requires the computation of Q or q along a reversible path between the states. The classical entropy relationships given by Equations 7.3 to 7.6 are in terms of the change in entropy. The absolute values of S or s may be found by using the reference S = s = 0 when T = 0 from Equation 7.1. At a temperature of absolute zero, only a single state is available to the system, P = 1, and consequently the reference of zero entropy at absolute zero is used to tabulate the values of specific entropy for gases. However, in the steam tables, the entropy of saturated liquid water at 0.01◦ C is assigned a zero value (see Tables A.3 and A.4 in Appendix A). For refrigerant R-134a, the entropy of saturated liquid at −40◦ C is taken as zero (Tables A.6 and A.7). As we are concerned with changes in entropy, the assignment of a reference state, where s = 0, is arbitrary. From a mathematical standpoint, it is interesting to note that in Equation 7.3, the factor, 1/T, converts the inexact differential, Q, into an exact differential, d S, for a reversible process.

7.3

The Clausius Inequality

Equation 7.2, which resulted from a Carnot cycle analysis, is true for any reversible cycle that is designed to receive heat Q H from a reservoir at TH and reject heat Q L to a reservoir at TL . We may express Equation 7.2 in a cyclic integral form    Q =0 (7.7) T rev for a reversible cycle. Now consider the irreversible cycle operating between the same reservoirs and assume that from the reservoir at TH , it receives the same heat, Q H , as the reversible cycle. Because the irreversible cycle is less efficient than the reversible cycle, it must reject more heat, that is, ( Q L ) irrev > ( Q L ) rev . This means that   QH QL − P2 > P1 . The diagram also identifies the specific entropies of saturated liquid, sf ,

188

Introduction to Thermal and Fluid Engineering T Critical Point

Pc

P3

Superheated Vapor Region

P2 P1

Subcooled Liquid Region

s

sf

sg sx sfg

s FIGURE 7.2 T-s diagram showing four pressures an pertinent entropy values.

saturated vapor, sg , superheated vapor, s, and the specific entropy at a particular quality, sx , all corresponding to a pressure, P1 . Tables A.3 to A.5 provide data for sf , sg , sfg = sg − sf and s. The specific entropy of a liquid-vapor mixture, sx , depends on the quality, x, and can be evaluated using Equation 4.7a or 4.7b.

Example 7.3 Consider the cycles described in Examples 7.1 and 7.2 and show them on a temperature-entropy diagram (Figure 7.3).

Solution We make the same assumptions that were made in Example 7.1 The cycle for Example 7.1 is shown in Figure 7.2 as 1-2-3-4-1. In process 1-2, a mixture of liquid and vapor at 20 kPa and a quality of x = 0.15 is compressed to saturated liquid T

2

10 MPa

x = 0.3572 1 x = 0.15

3

x = 0.6758

20 kPa 1'

4'

4

x = 0.90 s

FIGURE 7.3 T-s diagram for Example 7.3.

Entropy

189

water at 10 MPa. We read Table A.4 to find at 20 kPa sf = 0.8314 kJ/kg-K

and

sfg = 7.0779 kJ/kg-K

sf = 3.3580 kJ/kg-K

and sg = 5.6196 kJ/kg-K

and at 10 MPa

Then by Equation 4.7b at 20 kPa s1 = sf + xsfg = 0.8314 kJ/kg-K + (0.15)(7.0779 kJ/kg-K) = 1.8931 kJ/kg-K and at 10 MPa s2 = sf = 3.3580 kJ/kg-K Clearly, the entropy increases in process 1-2 and makes the process irreversible. In process 2-3, which occurs in the boiler, the heat is supplied isothermally at 10 MPa and changes the saturated liquid to saturated vapor. Since we have determined that the saturated vapor entropy value at 10 MPa is sg = 5.6196 kJ/kg-K, we can write s3 = sg = 5.6196 kJ/kg-K Although there is an increase in entropy from state 2 to state 3, this increase is associated with a reversible heat supply process. Process 3-4 occurs in the turbine where saturated vapor at 10 MPa expands to yield a mixture of liquid and vapor with a quality of x = 0.90. The entropy may be found via Equation 4.7b using values that have already been determined at 20 kPa s4 = sf + xsfg = 0.8314 kJ/kg-K + (0.90)(7.0779 kJ/kg-K) = 7.2015 kJ/kg-K Because s4 > s3 , the process of expansion in the turbine is irreversible. Process 4-1 occurs in the condenser where the heat is rejected isothermally at 333.1 K and the quality of the mixture of liquid and vapor decreases from x = 0.90 to x = 0.15. Observe that the decrease in entropy from state 4 to state 1 is due to a reversible heat rejection process. It may be noted that the net change in entropy over the cycle is zero and that because processes 1-2 and 3-4 are irreversible, the cycle is irreversible. The cycle for Example 7.2 is shown in Figure 7.2 as 1 -2-3-4 -1 and the entropies at state 1 and 4 may be determined as follows: s1 = sf + xsfg = 0.8314 kJ/kg-K + (0.3572)(7.0779 kJ/kg-K) = 3.3596 kJ/kg-K and s4 = sf + xsfg = 0.8314 kJ/kg-K + (0.6758)(7.0779 kJ/kg-K) = 5.6146 kJ/kg-K We note that s1 ≈ s2 and s4 ≈ s3 indicating that the compression from state 1 to state 2 and the expansion from state-3 to state-4 are isentropic (reversible adiabatic) processes. The cycle 1 -2-3-4 -1 is therefore reversible. Because entropy is a property of the system, its net change is zero.

Example 7.4 The compressor of a refrigeration unit receives saturated R-134a vapor at a

temperature of −20◦ C. The vapor is compressed and delivered to a condenser as a superheated vapor at 1.2 MPa (Figure 7.4). Assuming the compression process to be isentropic, determine (a) the temperature of the superheated vapor at the compressor outlet, (b) the

190

Introduction to Thermal and Fluid Engineering T 2

1.2 MPa

1

–20°C

s FIGURE 7.4 T-s diagram for the compressor in Example 7.4.

work input into the compressor, (c) the change in the specific internal energy of the refrigerant, and (d) the specific volume of the refrigerant at the compressor outlet.

Solution Assumptions and Specifications (1) The compression process is isentropic. (2) The system is the refrigerant in the compressor. (3) There are no potential and kinetic energy changes. (4) The system operates in the steady state. (a) The specific entropy of saturated R-134a vapor at −20◦ C (state 1) may be read from Table A.6 s1 = sg = 0.9332 kJ/kg-K Because the compression process is isentropic from state 1 to state 2 (superheated vapor at 1.2 MPa), there is no change in entropy so that s2 = s1 = 0.9332 kJ/kg-K When we consult Table A.8 at P = 1.2 MPa, we find that the value, s2 = 0.9332 kJ/kg-K, lies between the tabulated values of s = 0.9164 kJ/kg-K at 50◦ C and s = 0.9527 kJ/kg-K at 60◦ C. A linear interpolation with an interpolation factor, F , F =

0.9332 kJ/kg-K − 0.9164 kJ/kg-K = 0.46 0.9527 kJ/kg-K − 0.9164 kJ/kg-K

gives T = 50◦ C + (0.46)(60◦ C − 50◦ C) = 54.6◦ C ⇐ (b) For the compressor as a control volume with isentropic compression, the rate of ˙ cv = 0, and Equation 5.19 that ignores the changes in kinetic and potential heat transfer, Q energies gives the specific compressor work, wc as wc =

˙ cv W = h1 − h2 m ˙

Entropy

191

Noting that h 1 is equal to h g at −20◦ C, we have from Table A.6 h 1 = h g = 235.31 kJ/kg To find h 2 we interpolate using the data for h from Table A.8. Then h 2 = 275.52 kJ/kg + (0.46)(287.44 kJ/kg − 275.52 kJ/kg) = 281.00 kJ/kg and wc = h 1 − h 2 = 235.31 kJ/kg − 281.00 kJ/kg = −45.69 kJ/kg ⇐ This negative value shows that this is the work required to drive the compressor. (c) The change in specific internal energy may be obtained in similar fashion. Noting that u1 is equal to ug at −20◦ C, we have from Table A.6 u1 = ug = 215.84 kJ/kg To find u2 we interpolate using the data for u from Table A.8. u2 = 254.98 kJ/kg + (0.46)(265.42 kJ/kg − 254.98 kJ/kg) = 259.78 kJ/kg Then the change in specific internal energy is u2 − u1 = 259.78 kJ/kg − 215.84 kJ/kg = 43.94 kJ/kg ⇐ (d) The specific volume, v2 can be found by interpolating the data for v from Table A.8 v2 = 0.01712 m3 /kg + (0.46)(0.01835 m3 /kg − 0.01712 m3 /kg) or v2 = 0.01769 m3 /kg ⇐

7.5

The Gibbs Property Relations

Because entropy is not directly measurable, its numerical values appearing in Tables A.3 through A.8 and A.12 in Appendix A are calculated from a knowledge of measurable properties such as pressure, temperature, and specific volume and utilizing the thermodynamic relations that relate these properties to entropy. Two such relations are the Gibbs equations also called the TdS equations. Consider a closed stationary system composed of a simple compressible substance and experience heat and work interactions at its boundary. The differential form of the first law of thermodynamics for this system is Q − W = dU

(7.10)

If the heat and work transfers are reversible, then Equation 7.10 may be written as Qrev − Wrev = dU

(7.11)

Qrev = TdS

(7.12)

From Equation 7.3 it follows that

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Introduction to Thermal and Fluid Engineering

and, for a closed system, the only form of work that is reversible is the flow work (or PdV work) Wrev = PdV

(7.13)

When we substitute Equations 7.12 and 7.13 into Equation 7.11 we obtain, after a little rearrangement, the first Gibbs or TdS equation TdS = dU + PdV

(7.14)

Tds = du + Pdv

(7.15)

or

on a unit mass basis. Although Equations 7.14 and 7.15 are derived on the assumption of a reversible process, they are relations involving properties and, because changes in properties are independent of any particular process, must apply to irreversible processes as well. The second Gibbs or TdS relationship may be derived by introducing the expression for the enthalpy H = U + PV

(7.16)

Differentiation of Equation 7.16 provides dH = dU + PdV + VdP or dU + PdV = dH − VdP

(7.17)

When we replace dU + PdV on the right-hand side of Equation 7.14 with dH − VdP, Equation 7.14 may be expressed as TdS = dH − VdP

(7.18)

Tds = dh − vdP

(7.19)

or

on a unit mass basis. Equation 7.18 is the second Gibbs or TdS equation. Equations 7.15 and 7.19 are employed in the next section to develop expressions for the change in entropy of solids, liquids, and ideal gases in terms of pressure, temperature, and specific volume.

7.6

Entropy Change for Solids, Liquids, and Ideal Gases

The entropy change for a solid or liquid can be expressed in terms of a single property but the entropy change for an ideal gas requires a knowledge of two properties.

Entropy

193

7.6.1 Entropy Change for Solids and Liquids Consider Equation 7.15. For a solid or a liquid, the change in specific volume with temperature is rather negligible. Hence, with du = cdT ds =

du dT =c T T

(7.20)

If c¯ is the average value of c between the temperatures, T1 and T2 , Equation 7.20 may be integrated to give s2 − s1 = c¯ ln

T2 T1

(7.21)

Equation 7.21 shows that the change in entropy for a solid or liquid may be found from just the knowledge of the temperatures.

Example 7.5 Two aluminum blocks, one at a temperature of 80◦ C and the other at 40◦ C interact thermally until an equilibrium temperature, Te , is reached. Each block has a mass of 2 kg and an average specific heat of 903J/kg-K. Determine (a) the equilibrium temperature and (b) the change in entropy.

Solution Assumptions and Specifications (1) The system consists of the two blocks. (2) There is no heat interaction between the blocks and the surroundings. (a) Let m and c be the mass and average specific heat of each block with T1 and T2 taken as the initial temperatures of the hot and cold blocks. Because there are no heat or work interactions between the blocks and the surroundings, the first law energy balance is mc(T1 − Te ) = mc(Te − T2 ) or Te =

1 1 (T1 + T2 ) = (80◦ C + 40◦ C) = 60◦ C 2 2

(333 K) ⇐

(b) With T1 = 80◦ C (353 K) and T2 = 40◦ C (313 K) we can use Equation 7.21 to provide the change in entropy for each block. For block 1 (the hot block) S1 = m¯c ln

Te T1

= (2 kg)(903J/kg-K) ln

333 K 353 K

= (1806 J/K) ln(0.9433) = −105.34 J/K Similarly, for block 2 (the cold block) S2 = m¯c ln

Te T2

= (2 kg)(903 J/kg-K) ln

333 K 313 K

= (1806 J/K) ln(1.0639) = 111.86 J/K

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Introduction to Thermal and Fluid Engineering

The two blocks represent the system and the change in entropy will be S = S1 + S2 = −105.34 J/K + 111.86 J/K) = 6.52 J/K ⇐

7.6.2 Entropy Change for an Ideal Gas For an ideal gas, the change in specific volume cannot be neglected. Further, the specific heat is a function of temperature. The change in specific internal energy is du = c v (T)dT

(3.30a)

Equation 3.30a, along with the ideal gas law RT v

P= allows us to put Equation 7.15 into the form ds = c v (T)

dT dv +R T v

(7.22)

If the average value of c v between the temperatures T1 and T2 is designated as c¯ v , then Equation 7.22 may be integrated to yield s2 − s1 = c¯ v ln

T2 v2 + R ln T1 v1

(7.23)

Alternatively, we may employ Equation 7.19 to develop an expression for the change in entropy using the specific heat at constant pressure. Noting that dh = c p (T)dT

(3.30b)

Equation 3.30b, along with the ideal gas law, allows us to put Equation 7.19 into the form ds = c p (T)

dT dP −R T P

(7.24)

If the average value of c p between the temperatures T1 and T2 is designated as c¯ p , then an integration of Equation 7.24 yields s2 − s1 = c¯ p ln

T2 P2 − R ln T1 P1

(7.25)

We see from Equations 7.23 and 7.25 that, unlike a solid or a liquid, the change in entropy of an ideal gas requires a knowledge of two properties.

Example 7.6 Carbon dioxide, initially at a pressure of 100 kPa and a temperature of 300 K, is compressed to a pressure of 300 kPa and a temperature of 400 K. The mass of carbon dioxide is 0.20 kg and the compression work is 200 kJ. Determine (a) the heat transferred and (b) the change in entropy for this process.

Entropy

195

Solution Assumptions and Specifications (1) The carbon dioxide is an ideal gas and is the system. (2) The entire process operates under quasi-equilibrium conditions. (3) Kinetic and potential energy changes are negligible. (4) The work of compression is specified. Table A.1 gives R = 0.189 kJ/kg-K and Table A.2 may be employed to provide the average specific heats at 350 K c¯ v = 0.706 kJ/kg-K

and c¯ p = 0.895 kJ/kg-K

(a) The first law expression for this process is Q − W = U2 − U1 = m¯c v (T2 − T1 ) and we note that the work of compression specified as 200 kJ is negative because it is done on the carbon dioxide. Hence, Q = W + m¯c v (T2 − T1 ) = −200 kJ + (0.20 kg)(0.706 kJ/kg-K)(400 K − 300 K) = −200 kJ + 14.12 kJ = −185.88 kJ ⇐ The negative sign shows that heat is lost to the surroundings. (b) The change in entropy is evaluated using Equation 7.25 s2 − s1 = c¯ p ln

T2 P2 − R ln T1 P1

= (0.895 kJ/kg-K) ln

400 K 300 kPa − (0.189 kJ/kg-K) 300 K 100 kPa

= 0.2575 kJ/kg-K − 0.2076 kJ/kg-K = 0.0499 kJ/kg-K Then S = m(s2 − s1 ) = (0.20 kg)(0.0499 kJ/kg-K) = 0.0100 kJ/K ⇐

7.7

The Isentropic Process for an Ideal Gas

In an isentropic process, there is no change in entropy (S = 0). Thus, for an ideal gas in an isentropic process, Equation 7.23 gives s2 − s1 = c¯ v ln

T2 v2 + R ln =0 T1 v1

or ln

  R/¯cv T2 R V2 R V1 V1 = − ln = ln = ln T1 c¯ v V1 c¯ v V2 V2

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Introduction to Thermal and Fluid Engineering

and it then follows that T2 = T1



V1 V2

 R/¯cv

We note from Equation 3.35a that R =k−1 c¯ v which allows us to write T2 = T1



V1 V2

k−1 (7.26)

If we put s2 − s1 = 0 in Equation 7.25, a similar development brings us to ln

T2 R P2 = ln T1 c¯ p P1

It then follows that T2 = T1



P2 P1

 R/¯c p

and we note from Equation 3.35b that R k−1 = c¯ p k and hence, T2 = T1



P2 P1

(k−1)/k (7.27)

When we equate Equations 7.26 and 7.27 we obtain  k−1  (k−1)/k T2 V1 P2 = = T1 V2 P1 which gives P2 = P1



V1 V2

k (7.28)

Equations 7.26 through 7.28 are the P-V-T relationships for an ideal gas undergoing an isentropic process. We may write Equation 7.28 as P1 V1k = P2 V2k

(7.29)

or more generally P Vk = C

(a constant)

The work in an isentropic process will be  V2 P2 V2 − P1 V1 mR(T2 − T1 ) W= PdV = = 1−k 1−k V1

(7.30)

(7.31)

Entropy

197 500 kPa T ?

100 kPa

298 K

s FIGURE 7.5 T-s diagram for the compressor in Example 7.7.

Example 7.7 An air compressor draws air at 100 kPa and 298 K and delivers it at a pressure

of 500 kPa (Figure 7.5). Assuming the compression process to be isentropic with k = 1.4 and specific heats constant at 300 K, determine (a) the final specific volume, (b) the final temperature of the air, and (c) the work of compression.

Solution Assumptions and Specifications (1) The compressor is the control volume. (2) The air is an ideal gas and is the system. (3) The entire process operates under quasi-equilibrium conditions. (4) Kinetic and potential energy changes are negligible. (a) The initial specific volume of the air may be determined from the ideal gas law. With R = 0.287 kJ/kg-K and c p = 1.005 kJ/kg-K from Table A.1, we can write v1 =

RT1 (0.287 kJ/kg-K)(298 K) = = 0.8553 m3 /kg P1 100 kPa

The final specific volume may now be found from a rearrangement of Equation 7.28  v2 = v1

P1 P2

1/k

 = (0.8553 m /kg) 3

100 kPa 500 kPa

1/1.4 = 0.2709 m3 /kg ⇐

(b) To find the final temperature, we use Equation 7.27  T2 = T1

P2 P1

(k−1)/k

 = (298 K)

500 kPa 100 kPa

0.4/1.4 = 472 K ⇐

(c) The first law of thermodynamics for a control volume, in the absence of kinetic and potential energy changes, on a unit mass basis is q cv − wcv = h 2 − h 1

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Introduction to Thermal and Fluid Engineering

For an isentropic process, q cv = 0, which then gives wcv = h 1 − h 2 For an ideal gas with constant specific heats, h = c p T. Thus, wcv = h 1 − h 2 = c p (T1 − T2 ) = (1.005 kJ/kg-K)(298 K − 472 K) = −174.87 kJ/kg ⇐ We note that the negative sign indicates that work is being done on the system (the air).

7.8

Isentropic Efficiencies of Steady Flow Devices

From a power production or power consumption perspective, it is highly desirable to consider the use of steady flow devices such as turbines, pumps, and nozzles in an isentropic or reversible adiabatic process. In actual practice, however, a reversible adiabatic process cannot be realized because of the presence of irreversibilities such as friction and heat transfer through a finite temperature difference. The concept of isentropic efficiency is used to compare the actual performance of a steady flow device with its ideal performance under isentropic conditions. It is defined in such a manner that its numerical value is always less than unity. For a steam or gas turbine, we define the isentropic efficiency as the ratio of the actual enthalpy change, h 1 − h 2 , to the enthalpy change in an isentropic process, h 1 − h 2s s =

h1 − h2 h 1 − h 2s

(7.32)

We see that, neglecting kinetic and potential energy changes, the isentropic efficiency is the ratio of the actual specific work output to the isentropic work output and because h 1 − h 2 < h 1 − h 2s , the isentropic efficiency is always less than unity. The actual and 1 T

P1

2 P2

2s

s FIGURE 7.6 T-s diagram for a turbine with isentropic operation between points 1 and 2s and actual operation between points 1 and 2.

Entropy

199 450°C

T

1

8 MPa

30 kPa 2s x = 0.822

2

x = 0.90 s

FIGURE 7.7 T-s diagram for the turbine in Example 7.8. Point 2 is at a quality of 0.900 and it is shown in the example that point 2s is at 0.822.

isentropic processes for a steam turbine and gas turbine are shown in Figure 7.6.

Example 7.8 Steam at 8 MPa and 450◦ C expands in a steam turbine to a pressure of 30 kPa

with a quality of x = 0.90. The mass flow rate of steam is 25 kg/s. Determine (a) the power output and (b) the isentropic efficiency of the turbine.

Solution Assumptions and Specifications (1) The water-steam mixture in the turbine forms the system. (2) The turbine is the control volume. (3) Steady flow exists. (4) Kinetic and potential energy changes are negligible. (a) With reference to Figure 7.7, state 1 corresponds to 8 MPa and 450◦ C and Table A.5 reveals that h 1 = 3273.5 kJ/kg

and s1 = 6.5574 kJ/kg-K

State 2 is at 30 kPa and data from Table A.4 provides h f = 289.10 kJ/kg

and

h fg = 2336.3 kJ/kg

Then, after noting that the expansion from state 1 to state 2 is not isentropic, we find that h 2 = h f + xh fg = 289.10 kJ/kg + (0.90)(2336.3 kJ/kg) = 289.10 kJ/kg + 2102.7 kJ/kg = 2391.8 kJ/kg and ˙ turb = m(h W ˙ 1 − h2) = (25 kg/s)(3273.5 kJ/kg − 2391.8 kJ/kg) = 22.04 MW ⇐

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Introduction to Thermal and Fluid Engineering

(b) To find the enthalpy at state 2s (h 2s ), we first find the quality after an isentropic expansion. With s1 = s2s = 6.5574 kJ/kg-K, we obtain from Table A.4 at 30 kPa sf = 0.9433 kJ/kg-K

sfg = 6.8260 kJ/kg-K

and

so that via Equation 4.8 x2s =

s1 − sf 6.5574 kJ/kg-K − 0.9433 kJ/kg-K = = 0.823 sfg 6.8260 kJ/kg-K

and then h 2s = h f + x2s h fg = 289.10 kJ/kg + (0.823)(2336.3 kJ/kg) = 289.10 kJ/kg + 1922.8 kJ/kg = 2211.9 kJ/kg The isentropic efficiency may be found using Equation 7.32 s = =

h1 − h2 3273.5 kJ/kg − 2391.8 kJ/kg = h 1 − h 2s 3273.5 kJ/kg − 2211.9 kJ/kg 881.7 kJ/kg = 0.831 1061.6 kJ/kg

(83.1%) ⇐

For a compressor, a pump or a diffuser operating between state 1 and state 2, the isentropic efficiency is s =

h 2s − h 1 h2 − h1

(7.33)

and the actual and isentropic processes for a pump and a gas compressor are shown in Figure 7.8.

P2

T 2

P1

2s

1

s FIGURE 7.8 T-s diagram for a pump, compressor, or diffuser with ideal and nonideal entropy changes.

Entropy

201 T

2 2s

0.50 MPa

1

–20°C

s FIGURE 7.9 T-s diagram for the compressor in Example 7.9.

Example 7.9 A refrigeration compressor takes saturated R-134a vapor at −20◦ C and compresses it to a pressure of 0.50 MPa (Figure 7.9). If the isentropic efficiency is 0.85, determine (a) the specific work required to drive the compressor and (b) the temperature of the refrigerant after compression.

Solution Assumptions and Specifications

(1) The compressor is the control volume. (2) The refrigerant is the system. (3) Steady flow exists. (4) The system operates in the steady state. (5) Kinetic and potential energy changes are negligible. (a) Table A.6 provides at −20◦ C h 1 = h g = 235.31 kJ/kg

and s1 = sg = 0.9332 kJ/kg-K

Table A.8 shows that at 0.50 MPa, s = 0.9264 kJ/kg-K at 20◦ C and s = 0.9597 kJ/kg-K at 30◦ C. With s2s = s1 = 0.9332 kJ/kg-K, we find an interpolation fraction F =

0.9332 kJ/kg-K − 0.9264 kJ/kg-K 0.0068 kJ/kg-K = = 0.204 0.9597 kJ/kg-K − 0.9264 kJ/kg-K 0.0333 kJ/kg-K

so that h 2s = 260.34 + (0.204)(270.28 kJ/kg − 260.34 kJ/kg) = 260.34 kJ/kg + 2.03 kJ/kg = 262.37 kJ/kg

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Introduction to Thermal and Fluid Engineering

We may solve Equation 7.33 for h 2 h2 = h1 +

h 2s − h 1 s

= 235.31 kJ/kg +

262.37 kJ/kg − 235.31 kJ/kg 0.85

= 235.31 + 31.84 kJ/kg = 267.15 kJ/kg Then, the specific work to drive the compressor will be wcomp = h 2 − h 1 = 267.15 kJ/kg − 235.31 kJ/kg = 31.84 kJ/kg ⇐ (b) To find the final temperature, we use the data of Table A.8 with interpolation T2 = 20◦ C + (0.204)(30◦ C − 20◦ C) = 20◦ C + 2.0◦ C = 22◦ C ⇐ For a water- or air-cooled compressor, the ideal compression process is a reversible isothermal process and, therefore, a more appropriate measure of the efficiency is the ratio of the isothermal specific work to the actual specific work. The isothermal specific work may be obtained from a consideration of the steady flow energy equation which, in differential form (neglecting kinetic and potential energy changes), may be written for a reversible process as q rev − wrev = dh

(7.34)

Use of Equations 7.12 and 7.19 permits the representation q rev = Tds = dh − vdP so that Equation 7.34 becomes dh − vdP − wrev = dh or wrev = −vdP

(7.35)

Now, for an ideal gas, v = RT/P, so that wrev = −RT

dP P

(7.36)

Because T is constant during an isothermal compression, T1 = T2 = T, and an integration of Equation 7.36 will reveal wT=C = RT ln

P2 P1

(7.37)

where the minus sign may be omitted when it is agreed that W is the specific work input to the compressor. Thus, the efficiency of the compressor will be comp =

wT=C RT1 ln P2 /P1 = h2 − h1 c¯ p (T2 − T1 )

(7.38)

Entropy

203 T 2

600 kPa, 1550 K

PVn = Constant

1

100 kPa, 300 K

s FIGURE 7.10 T-s diagram for the compressor in Example 7.10.

Example 7.10 A compressor for a gas turbine (Figure 7.10) draws air at 100 kPa and 300 K and delivers it at a pressure of 600 kPa and 550 K. Assuming that the compression is in accordance with P V n = a constant, determine (a) the isothermal efficiency and (b) the polytropic exponent, n.

Solution Assumptions and Specifications (1) The compressor is the control volume. (2) The air is an ideal gas and is the system. (3) The system operates in the steady state. (4) Steady flow exists. (5) Kinetic and potential energy changes are negligible. Table A.1 gives R = 0.287 kJ/kg-K and with an average temperature of Tavg =

1 (300 K + 550 K) = 425 K 2

We find from Table A.2 (with interpolation) that c¯ p = 1.0165 kJ/kg-K (a) The thermal efficiency of the compressor may be found directly from Equation 7.38. With P2 /P1 = 600 kPa/100 kPa = 6, comp =

RT1 ln P2 /P1 (0.287 kJ/kg-K)(300 K) ln 6 = = 0.607 ⇐ c¯ p (T2 − T1 ) (1.0165 kJ/kg-K)(550 K − 300 K)

(b) The ideal gas law gives v1 =

RT1 (0.287 kJ/kg-K)(300 K) = = 0.861 m3 /kg P1 100 kN/m2

v2 =

RT2 (0.287 kJ/kg-K)(550 K) = = 0.263 m3 /kg P2 600 kN/m2

and

204

Introduction to Thermal and Fluid Engineering P1 = 600 kPa T T1 = 1200 K

1

P2 = 100 kPa T2 = 800 K

2 2s

s FIGURE 7.11 T-s diagram for the nozzle in Example 7.11.

and the polytropic exponent derives from P1 v1n = P2 v2n or n=

ln P2 /P1 ln (600 kPa/100 kPa) = = 1.51 ⇐ ln v1 /v2 ln ( 0.861 m3 /kg/0.263 m3 /kg)

We note that because n =  k, the process is not isentropic. Finally, we consider the isentropic efficiency of a nozzle that may be defined as the ratio of the actual specific kinetic energy at the nozzle exit to the specific kinetic energy at the exit of the nozzle obtained under isentropic expansion s =

Vˆ 2act Vˆ 2

(7.39)

S=C

Example 7.11 The nozzle of a jet engine (Figure 7.11) is fed with hot products of combustion at 600 kPa and 1200 K. The products exit the nozzle at 100 kPa and 800 K. Assuming the products of combustion to behave as an ideal gas with c¯ p = 1.060 kJ/kg-K and k = 1.37, determine (a) the actual nozzle exit velocity if the expansion is adiabatic, (b) the nozzle exit velocity if the expansion is isentropic, and (c) the isentropic efficiency of the nozzle.

Solution Assumptions and Specifications (1) The nozzle is the control volume. (2) The products of combustion are an ideal gas and comprise the system. (3) The system operates in the steady state. (4) Steady flow exists. (5) There are no heat or work interactions between the nozzle and the environment. (6) Kinetic energy at the inlet is negligible. (7) Potential energy changes are negligible. (8) The average specific heat is specified.

Entropy

205

(a) For the specified conditions, with subscripts 1 and 2 designating the inlet and outlet respectively, the energy equation becomes h1 = h2 +

Vˆ 22 2

or Vˆ 2 = [2(h 1 − h 2 )]1/2 = [2¯c p (T1 − T2 )]1/2 Hence, Vˆ 2 = [2(1.060 kJ/kg-K)(1000 J/kJ)(1200 K − 800 K)]1/2 = 920.9 m/s ⇐ (b) For isentropic flow, the temperature, T2s at the nozzle exit is given by Equation 7.27  T2s = T1

P2 P1

(k−1)/k

 = (1200 K)

100 kPa 600 kPa

0.37/1.37

= (1200 K)(0.1667) 0.270 = (1200 K)(0.6164) = 739.6 K The velocity, Vˆ 2s is then given by Vˆ 2s = [2¯c p (T1 − T2s )]1/2 = [2(1.06 kJ/kg-K)(1000 J/kJ)(1200 K − 739.6 K)]1/2 = 987.9 m/s ⇐ (c) The use of Equation 7.39 then gives the nozzle efficiency noz =

7.9

Vˆ 22 (920.9 m/s) 2 = = 0.869 (987.9 m/s) 2 Vˆ 22s

(86.9%) ⇐

The Entropy Balance Equation

Besides the mass and energy conservation equations that were treated in detail in Chapters 3 and 5, another equation of fundamental importance in thermodynamics is the entropy balance equation. In this section, we develop two steady-state entropy balance equations, one for the closed system or control mass and one for the open system or control volume. 7.9.1 The Entropy Balance for Closed Systems Figure 7.12 shows a closed system with heat and work interactions at its boundary. The system is assumed to be at the absolute temperature, T, and we will let the absolute temperature of the surroundings be fixed at To . Because of the heat transfer, Q, the closed system experiences an increase in entropy of Q/T. The internal irreversibilities within the closed system, if any, would give rise to entropy generation, Sgen , and because work transfer, W, is entropy free, we may write the entropy balance equation as Ssys =

Q + Sgen T

(7.40)

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Introduction to Thermal and Fluid Engineering Q

W FIGURE 7.12 A closed system with heat and work interactions at its boundary.

Now consider the surroundings that have experienced a decrease in entropy, So , given by So = −

Q To

(7.41)

The change in entropy of the universe (the system plus the surroundings), designated as Su , is then seen to be   1 1 Su = Q − + Sgen (7.42) T To The temperature of the surroundings, To , must be greater than the boundary temperature, T. For the heat transfer to occur from the surroundings to the system, and, because Sgen ≥ 0 (zero only if internal irreversibilities are absent), it follows from Equation 7.42 that Su ≥ 0

(7.43)

where the equality applies if Q is reversible, that is, if T = To and Sgen = 0. Equation 7.43 establishes the principle of increase of entropy, which states that the entropy of the universe always increases. For an adiabatic closed system, Q = 0, which makes So = 0, it follows from Equations 7.40 and 7.41 that Ssys = Su = Sgen

(7.44)

Example 7.12 Saturated steam at 200 kPa is contained in a closed, stationary vessel and is cooled by water surrounding the vessel. The mass of the vapor is 2 kg and during the period in which the steam condenses and is converted to saturated liquid water, the cooling water temperature rises from 25 to 35◦ C (Figure 7.13). Determine the net change in entropy for (a) the steam and (b) the cooling water.

Solution Assumptions and Specifications (1) The 2 kg of steam-water mixture is the system. (2) The system operates in the steady state. (3) There is enough cooling water to allow all of the steam to condense. (4) A stationary, closed, and rigid vessel is specified. (5) The temperature of the surroundings may be taken as the average between 35◦ C and 25◦ C or 30◦ C.

Entropy

207

Saturated Steam @ 200 kPa 2 kg Cooling Water 25°C

35°C

2 kg Condenses FIGURE 7.13 Configuration for Example 7.12.

The strategy here is to first determine the heat transferred across the boundaries of system. A stationary, closed, and rigid vessel is specified and for this closed system, the first law of thermodynamics in the absence of work is Q = U2 − U1 = mu Table A.4 may be read at 200 kPa to reveal that ufg = 2024.7 kJ/kg, which then gives Q = mufg = (2 kg)(2024.7 kJ/kg) = 4049.4 kJ The temperature of the system (the steam) is the saturation temperature corresponding to 200 kPa. Table A.4 provides T = 120.2◦ C (393.2 K). (a) The decrease in the entropy of the steam is S = −

Q 4049.4 kJ =− = −10.30 kJ/K ⇐ T 393.2 K

(b) The increase in the entropy of the cooling water at 30◦ C (303 K) is S =

Q 4049.4 kJ = = 13.36 kJ/K ⇐ T 303 K

We note that the net change in entropy is the algebraic sum of the entropy change of the water and the steam S = −10.30 kJ/K + 13.36 kJ/K = 3.06 kJ/K ⇐ 7.9.2 The Entropy Rate Balance for an Open System (Control Volume) The increase of entropy principle can be extended to an open system or control volume. Figure 7.14 shows a multiple input-multiple output control volume where a typical stream, i, with mass flow, m ˙ i , and specific entropy, si , is carrying entropy into the control volume at a rate of m ˙ i si . In a similar fashion, we may consider an exiting stream, e, with mass flow, m ˙ e , and specific entropy, se , carrying entropy out of the control volume at a rate of m ˙ e se . Summations over all inlet and outlet streams will then give the total flow of entropy into and out of the control volume m ˙ i si and m ˙ e se i

e

208

Introduction to Thermal and Fluid Engineering . M1, S1

Surface j

. Qj @ Tj

. M2, S2

. M3, S3

. M4, S4 . M5, S5 FIGURE 7.14 Multiple input-multiple output control volume.

Let the control volume consist of a number of isothermal surfaces. If surface, j, at an ˙ j , then the rate at which entropy absolute temperature, Tj , receives heat at the rate of Q ˙ enters the control volume through surface- j is Q j /Tj . The rate at which entropy enters the control volume over all of its surfaces is then Q ˙j Tj j If the control volume generates entropy at the rate of S˙ gen due to internal irreversibilities, the entropy rate balance for steady state and steady flow may be written as i

m ˙ i si +

Q ˙j + S˙ gen = m ˙ e se Tj e j

(7.45)

We see that with regard to a control volume, the rate at which entropy enters due to mass transfer at all of the inlet ports plus the rate at which entropy enters due to heat transfers at its confining surfaces plus any entropy generated due to internal irreversibilities is equal to the rate at which entropy leaves due to mass transfer at all of the outlet ports.

Example 7.13 Superheated steam enters a nozzle at 5 MPa and 300◦ C and exits the nozzle at 200 kPa and a quality of 0.95 (Figure 7.15). The rate of heat loss from the nozzle to the surroundings is 10 kW. Assume that the nozzle surface is at the average between the inlet and outlet steam temperatures. The mass flow rate is 2 kg/s. Determine the rate of entropy generation in the nozzle.

Solution Assumptions and Specifications (1) The 2 kg of steam-water mixture is the system. (2) The nozzle is the control volume.

Entropy

209 10 kW 5 MPa

2 kg/s

300°C

200 kPa x = 0.95

FIGURE 7.15 Steam flow through a nozzle for Example 7.13.

(3) The system operates in the steady state with steady flow. (4) The temperature of the nozzle wall may be taken as the average of the inlet and outlet stream temperatures. This is a single input-single output steady-state steady flow system with a single heat transfer from a uniform temperature surface. When we observe that the heat transferred is out of the system, we may write Equation 7.45 as ms ˙ i−

˙ Q + S˙ gen = ms ˙ e T

or ˙ Q ms ˙ i + S˙ gen = + ms ˙ e T where m ˙ is the mass flow rate of the steam. At 5 MPa and 300◦ C, Table A.5 provides si = 6.2085 kJ/kg-K and at the outlet of the nozzle at 200 kPa, Table A.4 gives sf = 1.5295 kJ/kg-K

and

sfg = 5.5974 kJ/kg-K

The entropy at the exit will then be se = sf + xsfg = 1.5295 kJ/kg-K + (0.95)(5.5974 kJ/kg-K) = 1.5295 kJ/kg-K + 5.3175 kJ/kg-K = 6.8470 kJ/kg-K At the exit of the nozzle where P = 200 kPa, Table A.4 reveals that Tsat = 120.2◦ C so that the average temperature of the nozzle walls will be T=

1 (300◦ C + 120.2◦ C) = 210.1◦ C 2

(483.1 K)

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Introduction to Thermal and Fluid Engineering

We may now solve for Sgen Sgen = =

˙ Q + m(s ˙ e − si ) T 10 kW + (2 kg/s)(6.8470 kJ/kg-K − 6.2085 kJ/kg-K) 483.1 K

= 0.0207 kW/K + (2 kg/s)(0.6385 kJ/kg-K) = 1.2977 kW/K

7.10

Summary

Entropy is a measure of molecular disorder. For our purposes, classical thermodynamics defines the difference in specific entropy as  s2 − s1 = 1

2

 q  T

rev

Entropy is a property and, in any process, is independent of the path. Absolute values are tabulated based on various substance standards. The Clausius inequality states that When a system operates in a cycle and the differential heat, Q, added or removed at a point is divided by the absolute temperature, T, at that point, the algebraic sum of all Q/T values is zero for a reversible cycle and less than zero for an irreversible cycle. The Gibbs property relations, or TdS equations, are TdS = dU + PdV

(7.14)

Tds = du + Pdv

(7.15)

TdS = dH − VdP

(7.18)

Tds = dh − vdP

(7.19)

and

For solids and liquids, the specific energy change is given by s2 − s1 = c¯ ln

T2 T1

(7.21)

where c¯ is the average specific heat. For ideal gases s2 − s1 = c¯ v ln

T2 v2 + R ln T1 v1

(7.23)

Entropy

211

and s2 − s1 = c¯ p ln

T2 P2 − R ln T1 P1

(7.24)

For an isentropic process of an ideal gas, the P-V-T relations are T2 = T1



V1 V2

k−1

T2 = T1



P2 P1

(k−1)/k

P2 = P1



V1 V2

k

and the work will be 

V2

W=

PdV =

V1

P2 V2 − P1 V1 mR(T2 − T1 ) = 1−k 1−k

(7.31)

Isentropic efficiencies of steady flow devices include the turbine s =

h1 − h2 h 1 − h 2s

(7.32)

s =

h 2s − h 1 h2 − h1

(7.33)

Vˆ 2act Vˆ 2

(7.39)

the compressor or pump

and the nozzle s =

S=C

The entropy balance equation for the closed system is Ssys =

Q + Sgen T

(7.40)

and for the open system or control volume i

7.11

m ˙ i si +

Q ˙j + S˙ gen = m ˙ e se Tj e j

(7.45)

Problems

The Clausius Inequality 7.1: The following heat engines operate between a hot reservoir at 1000 K and a cold reservoir at 300 K. Verify the validity of the Clausius inequality in (a) Q H = 500 kJ and Q L = 200 kJ, (b) Q H = 600 kJ and Q L = 180 kJ, and (c) Q H = 900 kJ and Q L = 250 kJ. 7.2: A heat engine draws heat from a flame at 800 K at the rate of 20 W. The engine rejects heat to an ocean at 290 K and the engine efficiency is 32%. Determine the cyclic integral  ˙  Q/T and comment on whether the engine satisfies the Clausius inequality. 7.3: A hot reservoir at 1500 K supplies 120 kJ to heat engine 1, which has a thermal efficiency of 28%. The heat rejected by engine 1 is supplied to engine 2, which operates with a thermal efficiency of 35%. Heat engine 2 rejects heat to a sink at 400 K and another

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Introduction to Thermal and Fluid Engineering 12 kJ to a sink at 300 K. Apply the Clausius inequality to determine whether engine 2 is reversible.

7.4: A heat engine receives 200 kJ from one source at 1000 K and 400 kJ from another source at 800 K. The engine rejects 120 KJ to a sink at 400 K and 120 kJ to another sink at 300 K. Apply the Clausius inequality to determine whether or not the engine is reversible. 7.5: Saturated liquid water at 5 MPa enters a boiler and is heated to saturated vapor at the same pressure. The vapor expands in an adiabatic turbine to 15 kPa with a quality of 0.85. The exhaust from the turbine enters a condenser where heat is extracted from the mixture and its quality is reduced to 0.20. The mixture than enters an adiabatic pump, which compresses it to saturated liquid at 5 MPa. Determine the value of the  ˙ cyclic integral  Q/T. 7.6: A heat engine draws 800 kJ from a hot reservoir at 600 K and rejects heat to a cold reservoir at 300 K. The work output from the engine is used to drive a heat pump, which draws 300 kJ from a cold reservoir at 280 K and delivers it to a hot reservoir at 310 K. The thermal efficiency of the engine is 25% and the coefficient of performance of the heat pump is 3.5. Determine whether the combination of the engine and the heat pump satisfies the Clausius inequality. 7.7: A vapor compression refrigeration cycle using R-134a executes four processes: Process 1-2 An isentropic compression, T1 = −8◦ C and s1 = 0.9066 kJ/kg-K. to T2 = 31.33◦ C and s2 = 0.9066 kJ/kg-K. Process 2-3 A constant pressure heat rejection to saturated liquid at P3 = 0.8 MPa. Process 3-4 An adiabatic throttling process to a mixture of liquid and vapor at P4 = 0.21704 MPa. Process 4-1 A constant pressure heat extraction from the evaporator.  ˙ Evaluate the cyclic integral  Q/T and comment on whether the system satisfies the Clausius inequality. Temperature-Entropy Diagrams 7.8: Determine the entropy of water for the following states and indicate the states on a T-s diagram, (a) T = 40◦ C and P = 5 MPa, (b) T = 45.80◦ C and P = 10 kPa, (c) T = 70◦ C and v = 4 m3 /kg, (d) T = 200◦ C, and (e) T = 325◦ C and P = 2.50 MPa. 7.9: Show the expansion of steam with T1 = 342.4◦ C and P1 = 15 MPa (saturated vapor) to T2 = 150◦ C and P2 = 200 kPa on a T-s diagram utilizing the entropy data for the three intermediate points, (1) T = 300◦ C and P = 8 MPa, (2) T = 250◦ C and P = 1.5 MPa, and (3) T = 200◦ C and P = 600 kPa. 7.10: Draw a T-s plot showing the throttling of saturated liquid water at 0.8 to 0.01 MPa. Indicate on your plot, the quality of the mixture corresponding to P = 0.60 MPa, P = 0.40 MPa, P = 0.20 MPa, and P = 0.0.01 MPa. 7.11: Determine the entropy of refrigerant R-134a for the following states and indicate the states on a T-S diagram: (a) T = 40◦ C and P = 0.09135 MPa, (b) T = −12◦ C and v = 0.08 m3 /kg, (c) P = −0.2 MPa and u = 170 kJ/kg, (d) P = 0.5 MPa and h = 275 kJ/kg, and (e) T = 40◦ C and v = 0.04633 m3 /kg.

Entropy

213

7.12: Refrigerant R-134a is compressed from T1 = −4◦ C (saturated vapor) to P2 = 1.0 MPa and T2 = 80◦ C. Show the path of compression on a T-s diagram indicating any three intermediate state points giving the values of T, s, and h. 7.13: Draw the cycle 1-2-3-4-1 on a T-s diagram with the state points in accordance with the following: State-1: T1 = −12◦ C, x1 = 0.90 State-2: T2 = 80◦ C, P2 = 1.0 MPa State-3: T2 = 35◦ C, P3 = 1.00 MPa State-4: T1 = −12◦ C, x4 = 0.20 7.14: A throttling of saturated refrigerant R-134a occurs from 6 to 0.6 bar. Draw a T-s plot marking the curve with quality values corresponding to P = 6, 4, 2, 1, and 0.6 bar. Entropy Changes for Solids, Liquids, and Ideal Gases 7.15: Two steel blocks, one at a temperature of 200◦ C and the other at 100◦ C, are brought into thermal contact with one another until thermal equilibrium is established. The mass of each block is 5 kg and the average specific heat of steel is 0.434 kJ/kg-K. Assuming no interaction with the surroundings, determine (a) the equilibrium temperature and (b) the change in entropy of the two blocks together as the system. 7.16: A 2-kg aluminum ball is removed from a furnace at 120◦ C. The ball is allowed to cool in an environment at 20◦ C until equilibrium is established. Assume the specific heat for aluminum to be 0.90 kJ/kg-K and determine (a) the change in entropy for the ball, (b) the change in entropy of the environment, and (c) the change of entropy of the ball and the environment taken together. 7.17: An ice cube of dimensions 2.5 cm × 2.5 cm × 2 cm at 0◦ C is placed in an insulated cup containing 0.4 kg of water at 16◦ C. Assuming that the final temperature is 0◦ C, determine (a) the amount of ice melted and (b) the change in entropy of the ice and water taken together. 7.18: A 1.5-kg copper block at 200◦ C is quenched in a water bath containing 6 kg of water at 20◦ C. Assume that the specific heats of copper and water are respectively 385J/kg-K and 4179J/kg-K with no heat loss to the surroundings and determine (a) the final equilibrium temperature, (b) the entropy change of the copper block, and (c) the entropy change of the water. 7.19: A hot block of mass, m1 , specific heat, c 1 , and a temperature of T1 is brought into thermal contact with a cold block of mass, m2 , specific heat, c 2 , and a temperature of T2 . If the heat loss to the surroundings is Q, show that the equilibrium temperature, Te , is given by Te =

m1 c 1 T1 + m2 c 2 T2 − Q m1 c 1 + m2 c 2

7.20: In a refrigeration process, 10 kg of saturated liquid water at 0◦ C is converted to ice at −5◦ C. Assume the latent heat of water to be 335 kJ/kg and c for ice to be 2040J/kg-K and determine the decrease of entropy of the water during the process. 7.21: Determine the change in entropy that occurs when 3 kg of oil (c = 1909J/kg-K) are heated from 20 to 50◦ C.

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7.22: A hot reservoir at 500 K is used to heat 12 kg of water from 25 to 85◦ C. Determine the change of entropy of the water and the hot reservoir using (a) c = 4178J/kg-K for water and (b) using the entropy data from Table A.3. 7.23: Using c p and c v values at an average temperature of 450 K, determine the change in entropy when 3 kg of air is heated at constant volume from P1 = 200 kPa and T1 = 300 K to P2 = 1000 kPa and T2 = 600 K. 7.24: A kilogram of nitrogen at 1200 kPa and 700 K expands to 200 kPa and 400 K. The work in the expansion process is 400 kJ. Use the specific heat values from Table A.1 to determine (a) the heat transfer and (b) the change in entropy of the nitrogen. 7.25: One kilogram of air at 900 kPa and 500 K expands to 100 kPa and 300 K. Determine the change in entropy of the air if (a) the air is cooled at 900 kPa from 600 to 300 K followed by an isothermal expansion from 900 to 100 kPa and (b) if the air is expanded to its final volume followed by a constant volume cooling to 300 K. 7.26: Consider the flow of 2 kg of helium at 300 kPa and 450 K through a throttling valve. The pressure of the helium after throttling is 100 kPa. Using specific heat values from Table A.1 and assuming them to be constant, determine the change in entropy. 7.27: Three kilograms of argon experience an increase in entropy of 0.65 kJ/K while executing an internally irreversible process. The initial temperature of the argon is 300 K. Using specific heat values from Table A.1 and assuming them to be constant, evaluate the final temperature of the argon for (a) a constant pressure process and (b) a constant volume process. 7.28: One kilogram of air executes a sequence of processes in a steady-flow system. Process 1-2

A polytropic compression (n = 1.2) from P1 = 100 kPa and T1 = 300 K to 600 kPa

Process 2-3

A constant pressure heat addition from T2 to T3 = 1200 K

Process 3-4

A polytropic expansion (n = 1.2) from P3 and T3 to P4 = 100 kPa

Process 4-1

A constant pressure heat rejection to T4 = 300 K

Assume that Table A.1 may be used for R and the specific heats and determine (a) the temperatures T3 and T4 and (b) the change in entropy for each of the processes. 7.29: Assuming 1 kg of an ideal gas with constant specific heats, determine the change in entropy between the initial and final states in each of the following cases: (a) Air; P1 = 120 kPa and T1 = 300 K; P2 = 480 kPa and T2 = 550 K (b) Argon; P1 = 100 kPa and T1 = 298 K; P2 = 350 kPa and v2 = 0.05 m3 /kg (c) Carbon dioxide; P1 = 110 kPa and T1 = 310 K; P2 = 420 kPa and v2 = 0.06 m3 /kg (d) Oxygen; P1 = 500 kPa and T1 = 500 K; P2 = 100 kPa and T2 = 325 K (e) Nitrogen; P1 = 800 kPa and T1 = 600 K; P2 = 200 kPa and T2 = 400 K 7.30: One kilogram of ethane at P1 = 100 kPa and T1 = 300 K is compressed to P2 = 600 kPa with s2 − s1 = −0.25 kJ/kg-K. Assuming ethane to be an ideal gas with properties given in Table A.1, determine (a) T2 and (b) v2 . 7.31: Nitrogen at 100 kPa and 300 K is contained in a piston-cylinder assembly. The mass is 0.6 kg and 250 kJ of heat is supplied to the nitrogen, The specific heats are constant as given in Table A.1. Determine for the nitrogen (a) the final temperature, (b) the final volume, and (c) the change in entropy.

Entropy

215

7.32: One kilogram of helium executes the following cycle in a closed system: Process 1-2

A constant volume compression from P1 = 100 kPa and T1 = 300 K to P2 = 300 kPa

Process 2-3

An isothermal expansion from P2 = 300 kPa to P3 = 100 kPa A constant pressure heat rejection to T1 = 300 K

Process 3-1

Sketch the cycle on a P-V diagram and, assuming that helium behaves as an ideal gas with constant specific heats given in Table A.1, determine (a) T2 , (b) v3 , (c) the heat transferred, (d) the work done, and (e) the change in entropy between state 3 and state 1. 7.33: A power cycle consists of three internally reversible processes: Process 1-2

A polytropic compression (n = 1.3) from P1 = 100 kPa and T1 = 300 K to P2 = 650 kPa

Process 2-3

An isothermal expansion from P2 = 650 kPa to P3 = 100 kPa

Process 3-1

A constant pressure compression back to P1 = 100 kPa

Show the cycle on a P-v diagram. Then, assume that the working fluid is 1 kg of nitrogen that may be taken as an ideal gas with specific heats taken as constant from Table A.1 and determine (a) T2 and v2 , (b) T3 and v3 , and (c) the changes in entropy in each of the three processes. Isentropic Relations for an Ideal Gas 7.34: Air is compressed isentropically from P1 = 100 Pa and T1 = 300 K to P2 = 500 kPa. Assuming air to be an ideal gas with constant specific heats, determine (a) T2 and (b) V1 /V2 . 7.35: An Otto cycle consists of four processes that are executed in a closed system: Process 1-2

An isentropic compression from T1 and v1 to T2 and v2

Process 2-3

A constant volume heat addition from T2 and v2 to T3 and v3 = v2

Process 3-4

An isentropic expansion from T3 and v3 to T4 and v4 = v1

Process 4-1

A constant volume heat rejection from T4 and v4 to T1 and v1

Assume that the air is an ideal gas with constant specific heats and draw the cycle on both P-v and T-s coordinates and show that the increase in entropy in process 2-3 is equal to the decrease in entropy in process 4-1. 7.36: For Problem 7.35, prove that T1 T4 = = T2 T3



v2 v1

k−1

7.37: Oxygen is compressed isentropically from P1 = 100 kPa and T1 = 398 K to P2 = 600 kPa. Assuming that oxygen is a perfect gas with constant specific heats, determine the specific work done on the oxygen when (a) the oxygen is compressed in a piston-cylinder arrangement and (b) when the oxygen is compressed in a steady-flow compressor. 7.38: One kilogram of air executes a sequence of processes in a steady-flow system. Process 1-2 An isentropic compression from P1 = 100 kPa and T1 = 300 K to P2 = 600 kPa

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Introduction to Thermal and Fluid Engineering Process 2-3 Process 3-4

A constant pressure heat addition from P2 and T2 to T3 = 1300 K An isentropic expansion from P3 and T3 to P4 = P1 and T4

Process 4-1 A constant pressure heat rejection from P3 and T3 to P4 and T4 Assume that air is an ideal gas with constant specific heats and determine (a) the temperatures T3 and T4 and (b) the changes in specific entropy, s3 − s2 and s1 − s4 . 7.39: One kilogram of methane executes a cycle shown in Figure P7.39. If P1 = 95 kPa, T1 = 298 K, and V1 /V2 = 4, assume that methane is an ideal gas with constant specific heats and determine (a) the pressure and temperature at state 2 and, if the pressure, P3 is 350 kPa, (b) the changes in specific entropy, s3 − s2 and s1 − s3 . P 2

P=C

3

PVk = C

PVk = C

1

P=C

4

V FIGURE 7P.39

7.40: Carbon monoxide undergoes a change if state from an initial temperature of 47◦ C and a pressure of 200 kPa to a final state of 107◦ C and a pressure of 475 kPa. Determine (a) the heat transferred and (b) the change in specific entropy. Isentropic Efficiencies of Steady-Flow Devices 7.41: A steam turbine receives steam at 6 MPa and 400◦ C and exhausts it to a condenser that operates at a pressure of 40 kPa. Kinetic and potential energy changes are negligible and the turbine has an isentropic efficiency of 0.80. Determine (a) the enthalpy and quality of the exhaust steam and (b) the specific work output from the turbine. 7.42: Superheated steam at 6 MPa and 800◦ C with a velocity of 15 m/s enters a steam turbine. The steam is exhausted at a pressure of 15 kPa with a velocity of 60 m/s and a quality of 0.95. The turbine produces 800 kW with a mass flow rate of 5000 kg/h. Determine (a) the rate of heat flow from the turbine to the surroundings and (b) the isentropic efficiency of the turbine. 7.43: Steam at 10 MPa and 650◦ C enters a turbine and leaves as saturated vapor at 10 kPa. Kinetic and potential energy changes are negligible. Determine (a) the specific work output of the turbine, (b) the change in specific entropy of the steam, and (c) the isentropic efficiency of the turbine. 7.44: A well-insulated steam turbine operating at steady state develops 8 MW of power for a steam flow of 10 kg/s. The steam enters at 4 MPa and leaves at 8 kPa with a quality of 0.92. Kinetic and potential energy changes are negligible. Determine the inlet temperature of the steam. 7.45: A gas turbine receives hot gases at 600 kPa and 1400 K and exhausts them to the atmosphere at 100 kPa. The isentropic efficiency is 0.88. Assume the hot gases to

Entropy

217

behave like an ideal gas with c p = 0.9861 kJ/kg-K and c v = 0.7472 kJ/kg-K and determine the temperature at the turbine exhaust. 7.46: A compressor draws in nitrogen at 100 kPa and 300 K and delivers it at 300 kPa and 500 K. Determine the isentropic efficiency assuming that nitrogen is an ideal gas with constant specific heats. 7.47: Air at 100 kPa and 27◦ C enters a compressor at the rate of 2 kg/s and leaves at 600 kPa. Assuming air to be an ideal gas with constant specific heats, plot (a) the temperature of the air leaving the compressor and (b) the power input to the compressor as a function of the isentropic efficiency of the compressor for values ranging from 0.50 to 1.00 in increments of 0.10. What conclusions can you draw from the plot? 7.48: A compressor for a gas turbine system draws air at 100 kPa and 298 K and delivers it at 700 kPa and 600 K. Determine (a) the polytropic exponent, n, if the compression follows P V n = constant and (b) the efficiency of the compressor. 7.49: A feedwater pump draws saturated liquid condensate at 10 kPa and delivers it to a boiler at 7.5 MPa. The isentropic efficiency of the pump is 0.87, and kinetic and potential energy changes are negligible. Determine the outlet enthalpy. 7.50: A compressor draws in refrigerant R-134a at 0.60 bar and −20◦ C and delivers it at 10 bar. Neglecting any heat interaction with the environment and any kinetic and potential energy changes, determine the exit temperature of the refrigerant and the specific work input to the compressor for (a) isentropic compression and (b) an 83% isentropic efficiency of the compressor. 7.51: Steam at 8 MPa and 550◦ C is expanded in a well-insulated and horizontal nozzle to saturated vapor at 90 kPa. Determine the nozzle exit velocity and the exit temperature for (a) an isentropic expansion and (b) a 76% isentropic efficiency of the nozzle. 7.52: Steam at 4 MPa and 450◦ C flows through a well-insulated and horizontal nozzle. The exit velocity is 800 m/s. The isentropic efficiency of the nozzle is 0.85. Determine the actual exit velocity if the inlet kinetic energy of the steam is negligible. The Entropy Balance Equation 7.53 Saturated liquid water at 150 kPa is contained in a piston-cylinder device where 950 kJ/kg is supplied to the water from a source at 200◦ C. Determine (a) the quality of the liquid-vapor mixture that is produced as a result of the heat supply, (b) the change in specific entropy of the water, and (c) the change in the specific entropy of the universe (water plus the source). 7.54: A piston-cylinder device contains a mixture of water and steam with a quality of 0.20 at 200 kPa. The energy supplied from the surroundings consists of 100 kJ/kg of heat from a source at 180◦ C and an unknown amount of specific work (kJ/kg). As a result of the energy supplied, the mixture is converted into saturated vapor. Determine (a) the amount of work supplied and (b) the entropy generated. 7.55: A closed stationary vessel contains 1.5 kg of saturated water vapor at 300 kPa. The vessel loses heat to a large body of coolant at 20◦ C. As a result of the heat loss, the vapor condenses and becomes a mixture with a quality of 0.40. Determine the change in entropy of the steam and the coolant. 7.56: One kilogram of refrigerant R-134a with a quality of 0.85 at 1.6 bar is contained in a piston-cylinder arrangement. Because of a heat loss of 300 kJ/kg to the surroundings, the quality of the mixture changes to 0.20. Determine the entropy generated in the cooling process.

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7.57: One kilogram of nitrogen is compressed adiabatically from 100 kPa and 300 K to 600 kPa. Because of internal irreversibilities, the work required for compression is 35% more than the work for isentropic compression. Assuming that the nitrogen is an ideal gas with constant specific heats, determine (a) the actual work required for compression, (b) the temperature of the nitrogen after compression, and (c) the entropy generated. 7.58: The inside and outside temperatures of a single pane window are 20◦ C and −5◦ C. The window glass is 4 mm thick with a surface area of 0.80 m2 and has a thermal conductivity of 1.4 W/m-K. Determine (a) the rate of heat conduction through the window and (b) the entropy generated in the window. 7.59: A plane wall of dimensions 1 m × 0.5 m × 0.01 m generates heat internally at a rate of 50 kW/m3 . The left face of the wall receives heat at the rate of 1.5 kW and is maintained at 40◦ C. The temperature on the right face of the wall is 80◦ C. Determine the rate of entropy generation in the wall. 7.60: Steam at 8 MPa and 450◦ C expands in a turbine to 50 kPa and a quality of 0.90. The outer surface of the turbine, which is at 85◦ C, loses heat to the surroundings at 85◦ C at the rate of 25 kW and the mass flow rate of the steam is 50 kg/s. Assuming that kinetic and energy changes are negligible, determine the rate of entropy generation in the turbine. 7.61: Refrigerant R-134a in the form of a saturated liquid at 8 bar flows at 1.2 kg/s through a throttling valve and expands to a pressure of 1.8 bar. The throttling valve receives 100 W from the surroundings at 25◦ C through its surface, which is at 18◦ C. Assuming that kinetic and energy changes are negligible, determine the rate of entropy generation during the process. 7.62: Nitrogen gas at 100 kPa and 300 K flows through a steady-flow compressor at the rate of 10 kg/s. The power input to the compressor is 500 kW but 100 kW is lost in the form of heat transfer through the casing to the surroundings. The pressure of the compressed nitrogen is 300 kPa. Assuming that nitrogen is an ideal as with c p = 1.044 kJ/kg-K and c v = 0.747 kJ/kg-K as the average values and 60◦ C as the average surface temperature of the compressor, determine the rate of entropy generation during the process. 7.63: Refrigerant R-134a enters a horizontal pipe of 15 cm inside diameter as a mixture of liquid and vapor with a quality of 0.20 at a pressure of 4 bar and a velocity of 8 m/s. The refrigerant leaves the pipe as saturated vapor at 2 bar. During the process, heat is transferred from the surroundings to the pipe, which is at an average temperature of −2◦ C. For steady flow, determine (a) the mass flow rate, (b) the exit velocity, (c) the rate of heat transfer from the surroundings, and (d) the rate of entropy generation in the pipe. 7.64: In a double-pipe heat exchanger, air at 3 kg/s enters the inner tube at 120 kPa and 320 K and leaves at 100 kPa and 270 K. The outer tube carries refrigerant R-134a, which enters as saturated liquid at 140 kPa and exits as saturated vapor at the same pressure. Heat transfer between the heat exchanger and the surroundings is negligible. For steady operation and neglecting kinetic and potential energy changes, determine (a) the mass flow rate of R-134a and (b) the rate of entropy production in the heat exchanger.

8 Gas Power Systems

Chapter Objectives •

To establish the basis for the consideration of gas power systems.

•

To describe the operation of the spark-ignition (Otto)1 and compression-ignition (Diesel)2 engines, the gas turbine or Brayton engine, and the jet engine.

To define what is meant by the air standard and cold air standard cycles. • To consider both the air standard and cold air standard performance of the ideal Otto, Diesel, Brayton or gas turbine, and jet engine cycles. •

To discuss the compression ratio, its limitations, and its effect on the performance of the Otto and Diesel cycles. • To show the effect of regeneration on the ideal Brayton cycle. •

8.1

Introduction

In this chapter, we focus our attention on gas power systems that always use a gas as a working fluid. The gas power systems that we will consider are the reciprocating internal combustion engine (both spark-ignition or the Otto cycle and compression-ignition or the Diesel cycle), the rotating gas turbine, and the aircraft jet engine. These power systems are open systems because the working fluid is exhausted at the end of each cycle. Section 3.6.3 was devoted to an introduction to cycle analysis. There we saw that the work done by a power cycle is equal to the heat input less the heat rejected Wcycle = Qin − Qout and that the thermal efficiency of the cycle is =

1 Nikolaus

Wcycle Qin − Qout Qout = =1− Qin Qin Qin

(3.18)

Otto (1832–1891) was a German inventor who built the first four-stroke cycle internal combustion

engine. 2 Rudolf

Diesel (1858–1913) was a German inventor who designed and built the Diesel engine.

219

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Introduction to Thermal and Fluid Engineering

These equations are applied to the ideal internal combustion and gas turbine cycles that are characterized by all of the following: 1. Operation is frictionless with reversible compressions and expansions. 2. All processes are quasi-static or quasi-equilibrium processes. 3. There is no heat transfer in the conduits connecting the component parts of the system. 4. The working substance is air that is treated as an ideal gas. Because air is assumed as the working fluid, in the analysis of gas power systems, considerable simplification is evident. Air standard analysis considers that 1. All processes are internally reversible. 2. A fixed amount of air is used. 3. The combustion process is represented as a heat addition from an external source. 4. The exhaust process is represented as a heat rejection to an external sink. In air standard analysis, variable specific heats are employed and we obtain specific internal energy and enthalpy values from the “air tables,” which are provided in Table A.12 in Appendix A. Because both the compression and expansion processes in the cycle are isentropic, use of the reference volume vri Vi ≡ vr j Vj and the reference pressure pri Pi ≡ pr j Pj are required. They will be considered in further detail as we proceed. If a cold air standard analysis is to be conducted, then the specific heats are taken as constants.

8.2

The Internal Combustion Engine

A single piston-cylinder assembly of an internal combustion engine is shown in Figure 8.1(a). The inside diameter of the cylinder is called its bore and the distance that the piston travels from bottom dead center to top dead center is called the stroke. The volume in the cylinder when the piston is at top dead center is called the clearance volume and the volume that is displaced by the piston in one stroke as it moves from bottom dead center to top dead center is called the displacement volume. The ratio of the cylinder volume at bottom dead center to the volume at top dead center taken respectively as V1 and V2 is called the compression ratio, r r≡

V1 V2

Internal combustion engines can operate on either the four-stroke or two-stroke cycle. The four-stroke cycle consists of four separate and distinct strokes and we employ Figure 8.1(b) to describe this cycle.

Gas Power Systems

221 Spark Plug or Fuel Injector

Valve

Valve

Clearance Volume

Top Dead Center

P

Bore w Po er

Stroke n sio es pr m Co

Bottom Dead Center

Piston

Exhaust Intake Bottom V Dead Center

Top Dead Center Reciprocating Motion (a)

(b)

FIGURE 8.1 (a) A piston-cylinder assembly in an internal combustion engine and (b) its operation.

1. The intake stroke begins with the piston at top dead center. The intake valve opens and for the spark-ignition engine, a charge of air and fuel (such as gasoline) is drawn into the cylinder. In the compression-ignition engine, the charge is air without the fuel. 2. The compression stroke begins with the piston at bottom dead center and ends with the piston at top dead center. During the compression stroke both intake and exhaust (exit) valves are closed, and during this stroke, the pressure and temperature of the charge are markedly increased. P

T 3 4

3 2

2 4 1

1 s

V (a)

(b)

FIGURE 8.2 The ideal air standard gasoline engine or Otto cycle, which operates on a four-stroke cycle: (a) the P-v plane and (b) the T-s plane.

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3. The power stroke begins with the piston at top dead center. This is the stroke in which a combustion process occurs with the valves shut. The combustion process is induced by a spark in the spark-ignition engine and by the injection of the fuel in the compression-ignition engine; the power is derived from the expansion of the air fuel mixture as the piston moves from top dead center to bottom dead center. 4. In the exhaust stroke, in which the exhaust valve is open, the spent air fuel mixture is discharged as the piston moves from bottom dead center to top dead center. At the conclusion of the exhaust stroke, the piston-cylinder assembly is ready to begin a new cycle. We note that two revolutions of the crankshaft are needed to execute a four-stroke cycle. The two-stroke cycle combines the intake and the compression strokes and the power and exhaust strokes by using intake and exit ports. Only one revolution is required to execute the two-stroke cycle and, because of the unavoidable mixing of the incoming air-fuel mixture and the exhaust gases, an engine using the two-stroke cycle is generally less efficient than one that employs the four-stroke cycle. We now turn our attention to the air standard Otto cycle.

8.3

The Air Standard Otto Cycle

8.3.1 Performance The performance of the automobile engine may be modeled by the ideal air standard Otto cycle with the T-s and P-v representations shown in Figure 8.2. The four strokes in the cycle may be summarized by •

Process 1-2: Isentropic compression

•

Process 2-3: Constant volume heat addition

•

Process 3-4: Isentropic expansion

•

Process 4-1: Constant volume heat rejection

We note that the compression and expansion processes are both isentropic and that the heat addition and exhaust processes occur at constant volume, and in accordance with its definition, the compression ratio is r≡

V1 V4 ≡ V2 V3

(8.1)

If all of the work and heat interactions are considered as positive quantities, we can represent the work of compression and expansion by the closed-system expressions W12 = m(u2 − u1 )

and

W34 = m(u3 − u4 )

(8.2)

respectively and the heat transfer during the power stroke by Q23 = m(u3 − u2 )

(8.3a)

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223

and the exhaust stroke by Q41 = m(u4 − u1 )

(8.3b)

We are quick to note that, because of the convention adopted here that all of the work and heat interactions are to be considered positive, W12 is the work input during compression and Q41 is the heat rejected by the system. Both of these quantities are represented by a positive number. The net work and heat of the cycle are evaluated as Wcycle = W34 − W12 = m[(u3 − u4 ) − (u2 − u1 )]

(8.4a)

Qcycle = Q23 − Q41 = m[(u3 − u2 ) − (u4 − u1 )]

(8.4b)

and

The cycle efficiency is the net work of the cycle divided by the heat input =

Wcycle m[(u3 − u4 ) − (u2 − u1 )] u4 − u1 = =1− Qin m(u3 − u2 ) u3 − u2

(8.5)

The specific internal energy values required by Equations 8.4 and 8.5 are functions of temperature and, in air standard analysis, use is made of the reference volumes that apply to the isentropic processes in the cycle   V2 vr 1 vr 2 = vr 1 = (8.6a) V1 r and

 vr 4 = vr 3

V4 V3

 = r vr 3

(8.6b)

These reference volumes are functions of temperature and may be found in the “air tables” provided. If the analysis is based on the cold air standard where the specific heat at constant volume, c v , is taken as constant, then Wcycle = mc v [(T3 − T4 ) − (T2 − T1 )]

(8.7a)

Qcycle = mc v [(T3 − T2 ) − (T4 − T1 )]

(8.7b)

and =

Wcycle mc v [(T3 − T4 ) − (T2 − T1 )] T4 − T1 = =1− Qin mc v (T3 − T2 ) T3 − T2

(8.7c)

In the cold air standard cycle, the temperature-volume relationships for the isentropic compression and expansion were developed in Chapter 7. We have seen that these relationships employ the specific heat ratio which, for air, is taken as k = c p /c v = 1.40. Hence, as indicated in Chapter 7 T2 = T1



V1 V2

k−1 = r k−1

(8.8a)

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Introduction to Thermal and Fluid Engineering

and T4 = T3



V3 V4

k−1 =

1 r k−1

(8.8b)

and Equations 8.8a and 8.8b show that T2 T3 = = r k−1 T1 T4

(8.8c)

The efficiency relationship of Equation 8.7c can be modified to   T4 − T1 T1 (T4 /T1 ) − 1 =1− =1− T3 − T2 T2 (T3 /T2 ) − 1 and using Equation 8.8c, we see that this can be adjusted to =1−

1 r k−1

(8.9)

The mean effective pressure (mep) is defined as the ratio of the net work of the cycle to the displacement volume, V1 − V2 mep =

Wcycle r Wcycle = V1 − V2 V1 (r − 1)

(8.10)

If the pressure on the piston during the entire power stroke is constant and equal to the mean effective pressure, then Wcycle = (mep)(displacement volume) or Wcycle = (mep)(stroke)((piston area) This shows that the mean effective pressure is a yardstick that provides a comparison of engines running at the same speed. We now turn to two examples concerning the Otto cycle. It is the intent of these examples to consider both the cold air standard performance and the air standard cycle performance, showing the differences and discrepancies between the pressures and temperatures at various points in the cycle as well as the efficiencies.

Example 8.1 In an Otto cycle to be analyzed on the cold air standard, the initial pressure is 0.92 bar, the initial temperature is 27◦ C and the bottom dead center volume of the pistoncylinder assemblies is 555 cm3 . The compression ratio is 7.75 and the maximum temperature during the cycle is not to exceed 2100 K. Determine (a) the pressure and temperature at each of the four points (as shown in Figure 8.3) of the cycle, (b) the thermal efficiency, and (c) the mean effective pressure.

Solution Assumptions and Specifications (1) The air contained in the system is an ideal gas. (2) The cycle consists of two constant volume and two isentropic processes. (3) All processes are quasi-static.

Gas Power Systems

225 P

2100 K

3

4

0.92 Bar, 27°C

2 555 cm3

1

V2

V

V1

FIGURE 8.3 P-V diagram for the Otto cycle in Example 8.1.

(4) There are no kinetic and potential energy changes. (5) A cold air standard analysis is specified. (a) For the conditions at the end of the isentropic compression (point 2 in Figure 8.3), we use Equation 8.8a with k = 1.40 and T1 = 300 K T2 = T1



V1 V2

k−1 = r k−1

T2 = T1r k−1 = (300 K)(7.75) 0.40 = 680.5 K ⇐ Then, because for an ideal gas, P1 V1 /T1 = P2 V2 /T2  P2 = P1

V1 V2



T2 T1



 = P1r

T2 T1



we have  P2 = (0.92 bar)(7.75)

680.5 K 300 K

 = 16.17 bar ⇐

For the conditions at the end of the constant volume heat addition (point 3 in Figure 8.3) the maximum temperature allowed in the cycle is specified T3 = 2100 K and the pressure will be in accordance with Equation 3.37  P3 = P2

T3 T2



 = (16.17 bar)

2100 K 680.5 K

 = 49.91 bar ⇐

For conditions at the end of isentropic expansion (point 4 in Figure 8.3) use Equation 8.8b T4 = T3



V3 V4

k−1 =

 k−1 1 r

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Introduction to Thermal and Fluid Engineering

so that with T3 = 2100 K



T4 = (2100 K)

1 7.75

0.40 = (2100 K)(0.441) = 925.8 K ⇐

Then, because P4 V4 /T4 = P3 V3 /T3



P4 = P3

V3 V4

we have 49.91bar P4 = 7.75





T4 T3



925.8 K 2100 K

P3 = r



T4 T3



 = 2.84 bar ⇐

(b) We obtain the thermal efficiency from Equation 8.9 =1−

1 r k−1

=1−

1 = 1 − 0.441 = 0.559 ⇐ (7.75) 0.40

(c) One of the inputs to Equation 8.10 for the mean effective pressure is the net work of the cycle, which, in turn, requires that we evaluate the mass in the closed system. For air, Table A.1 provides R = 287 J/kg-K

and c v = 0.718 kJ/kg-K

and the ideal gas law may then be employed to obtain the mass m=

P1 V1 (0.92 × 105 N/m2 )(5.55 × 10−4 m3 ) = = 5.930 × 10−4 kg RT1 (287 J/kg-K)(300 K)

We then use Equation 8.7a to determine the net work Wcycle = mc v [(T3 − T4 ) − (T2 − T1 )] = (5.930 × 10−4 kg)(0.718 kJ/kg-K) [(2100 K − 925.8 K) − (680.5 K − 300 K)] = (4.258 × 10−4 kJ/K)(1174.2 K − 380.5 K) = (4.258 × 10−4 kJ/K)(793.7 K) = 0.338 kJ Now, Equation 8.10 gives the mean effective pressure mep =

r Wcycle V1 (r − 1)

or mep =

(7.75)(0.338 kJ) = 699.1 kN/m2 (555 × 10−6 m3 )(6.75)

or

6.99 bar ⇐

Example 8.2 Repeat Example 8.1 using an air standard analysis. Solution We make the same assumptions as were made in Example 8.1 except that an air standard analysis is specified.

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227

At 300 K, Table A.12 gives vr 1 = 621.2 and u1 = 214.07 J/kg Then, to find the conditions at the end of the isentropic compression (point 2 in Figure 8.3) we use the reference value ratio vr 2 /vr 1 = V2 /V1 so that   V2 vr 1 621.2 vr 2 = vr 1 = = = 80.16 V1 r 7.75 We then use Table A.12 once again but this time with interpolation to find T2 and u2 T2 = 665.3 K ⇐

and u2 = 484.70 kJ/kg

Because for the ideal gas, P2 V2 /T2 = P1 V1 /T1 we see that      V1 T2 T2 P2 = P1 = P1r V2 T1 T1 or

 P2 = (0.92 bar)(7.75)

665.3 K 300 K

 = 15.81 bar ⇐

At the end of the combustion process (the constant volume heat addition) at point 3 in Figure 8.3, T3 is the maximum temperature permitted in the cycle and is specified at T3 = 2100 K ⇐ Table A.12 then gives vr 3 = 2.356 and

u3 = 1775.3 kJ/kg

and because the combustion process takes place at constant volume, Equation 3.37 gives us     T3 2100 K P3 = P2 = (15.81 bar) = 49.90 bar ⇐ T2 665.3 K For conditions at point 4 (see Figure 8.3 at the end of the isentropic expansion) vr 4 V4 = =r vr 3 V3 so that vr 4 = vr 3r = (2.356)(7.75) = 18.26 Then interpolation of Table A.12 gives T4 = 1112.6 K ⇐

and u4 = 856.33 kJ/kg

Finally, because we are working with an ideal gas, P4 V4 /T4 = P3 V3 /T3 and the value of P4 will be        V3 T4 P3 T4 49.90 bar 1112.6 K = = P4 = P3 = 3.41 bar ⇐ V4 T3 r T3 7.75 2100 K

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Introduction to Thermal and Fluid Engineering

Equation 8.9 applies only to the cold air standard. Instead, use Equation 8.5  = 1−

u4 − u1 u3 − u2

= 1−

856.33 kg/s − 214.07 kg/s 1775.3 kg/s − 484.70 kg/s

= 1−

642.26 kg/s 1290.6 kg/s

= 1 − 0.498 = 0.502 ⇐ (c) The mean effective pressure is obtained from Equation 8.10. The value of the mass can be obtained from the specified conditions at point 1 and will be equal to the value obtained in Example 8.1. Hence, m = 5.930 × 10−4 kg and we may find Wcycle , which Equation 8.10 requires, from Equation 8.4a. Wcycle = m[(u3 − u4 ) − (u2 − u1 )] = (5.930 × 10−4 m)[(1775.3 kJ/kg − 856.33 kJ/kg − (484.70 kJ/kg − 214.07 kJ/kg)] = (5.930 × 10−4 m)(918.97 kJ/kg − 270.63 kJ/kg) = (5.930 × 10−4 m)(648.34 kJ/kg) = 0.384 kJ Now Equation 8.10 gives us mep = =

r Wcycle V1 (r − 1) (7.75)(0.384 kJ) (555 × 10−6 m3 )(6.75)

= 794.4 kN/m2

or

7.94 bar ⇐

The goal of Examples 8.1 and 8.2 was to provide a means for the assessment of any discrepancies in the results of the two analysis procedures. The examples were contrived to provide comparison of the temperatures and pressures at each point in the cycle, the efficiencies and the mean effective pressures. The pressures and temperatures at each point in the cycle are tabulated in Table 8.1 for both the cold air standard and the air standard analyses. With regard to both the temperature and pressure, we note a slight discrepancy at point 2, no discrepancy at points 1 and 3 (point 1 is the specified cycle starting point), and a rather marked discrepancy at point 4. The error involved at point 4, using the more correct air standard value as a basis, amounts to about 17% in both temperature and pressure. We also see a discrepancy between the two values of the efficiency and the mean effective pressure. For the efficiency, the comparison is 0.559 for the cold air standard and 0.502 for the air standard. This represents an error of about 10%. For the mean effective pressure, the value for the cold air standard is 6.99 bar and for the air standard, it is 7.94 bar, an error of about 13%.

Gas Power Systems

229 TABLE 8.1

A Comparison of Temperatures and Pressures between Examples 8.1 and 8.2. The Cycle Points Correspond to Figure 8.3 Cycle Point

Cold Air Standard, T K

Air Standard, T K

Cold Air Standard, P bar

Air Standard, P bar

1 2 3 4

300 680.5 2100 925.8

300 665.3 2100 1112.6

0.92 16.17 49.91 2.84

0.92 15.81 49.90 3.41

Because the specific heats for air do indeed vary with temperature, the air standard analysis is presumed to give more accurate results than its cold air standard counterpart. Use of the cold air standard temperatures and pressures can have an impact on the actual design of an engine working on the Otto cycle and the cold air standard efficiency and mean effective pressure can have a bearing on performance comparisons with other engines. 8.3.2 The Compression Ratio and Its Effect on Performance We may refer again to Equation 8.9 =1−

1 r k−1

(8.9)

Thermal Efficiency, η

which is plotted in Figure 8.4 and which shows that, on a cold air standard basis, the thermal efficiency is directly proportional to the compression ratio. We find it tempting to think that the thermal efficiency of the Otto cycle has an upper limit of 100% as the compression ratio increases without bound. However, it is a fact that limitations on the compression ratio are made by the fuel itself. In the spark-ignition engine, the upper limit on the compression ratio is dictated by the ignition temperature of the fuel that must be below the temperature at the end of the isentropic compression. Otherwise a high-velocity, high-pressure movement of the flame front in the cylinder will propagate before the spark ignition. This causes autoignition, which manifests itself in knocking or pinging. These annoying phenomena are more apt to occur with low octane fuels and at compression ratios greater than 10.

Compression Ratio, r FIGURE 8.4 The cold air standard thermal efficiency of the Otto cycle as a function of the compression ratio.

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Introduction to Thermal and Fluid Engineering

We turn now to Design Example 2, which involves the establishment of the bore and stroke of an internal combustion engine to operate on the Otto cycle.

8.4

Design Example 2

Determine the bore and stroke in an Otto engine to operate on the air standard in an ideal Otto cycle. The intake pressure is 0.96 bar at 27◦ C with an air mass of 3.75 × 10−4 kg. The compression ratio is limited to 8.00 in order to circumvent any fuel detonation problems and the maximum temperature in the cycle is limited to 2250 K because of metallurgical reasons. The thermal efficiency of the cycle is to be  = 0.500 and the length of the stroke is to be four times the bore.

Solution Assumptions and Specifications

(1) The air contained in the system is an ideal gas. (2) The cycle consists of two constant volume and two isentropic processes. (3) All processes are quasi-static. (4) There are no kinetic and potential energy changes. (5) An air standard analysis is specified. Figure 8.5 shows the P-v diagram for the cycle. A trial-and-error procedure will be necessary and Table 8.2 summarizes the pertinent calculations at assumed compression ratios of 7.00, 7.50, 7.75, and 8.00. A plot of the cycle efficiency as a function of the compression ratio is provided in Figure 8.6 where it is observed that an efficiency of 0.500 is achieved when the compression ratio, r, is 7.83. The volume at point 1 in Figure 8.5 may be determined from the ideal gas law. We read from Table A.1 R = 287 J/kg-K P

3

4 2

P1 = 0.96 Bar T1 = 27°C m1 = 3.75 × 10–4 kg

1 V FIGURE 8.5 P-v diagram for the Otto cycle in Design Example 2.

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231

TABLE 8.2

Computations for Design Example 2 Assume r vr 1 at 300 K u1 at 300 K, kJ/kg vr 2 = vr 1 /r T2 at vr 2 K u2 at T2 , K T2 /T1 P2 = P1 r (T2 /T1 ), MPa T3 , K vr 3 at T3 u3 at T3 , kJ/kg T3 /T2 P3 = P2 (T3 /T2 ), MPa vr 4 = r vr 3 T4 , K u4 at vr 4 , kJ/kg u4 − u1 , kJ/kg u3 − u2 , kJ/kg  = (u4 − u1 )/(u3 − u2 ) =1−

7.00

7.50

7.75

8.00

621.2 214.07 88.743 640.4 466.07 2.135 14.347 2250 1.864 1921.3 3.512 50.39 13.048 1240.7 969.54 755.47 1455.23 0.519 0.481

621.2 214.07 82.837 657.3 478.91 2.191 15.775 2250 1.864 1921.3 3.423 54.00 13.980 1213.2 945.37 731.30 1442.39 0.507 0.493

621.2 214.07 80.155 665.3 485.14 2.218 16.502 2250 1.864 1921.3 3.382 55.81 14.440 1200.7 933.95 719.88 1436.16 0.501 0.499

621.2 214.07 77.650 673.1 491.28 2.244 17.231 2250 1.864 1921.3 3.342 57.60 14.912 1188.5 923.01 708.94 1430.02 0.496 0.504

so that at T = 27◦ C (300 K) V1 =

mRT ˙ (3.75 × 10−4 kg)(287J/kg-K)(300 K) 1 = = 3.363 × 10−4 m3 P1 0.96 × 105 Pa

Then with V2 =

V1 3.363 × 10−4 m3 = = 4.295 × 10−5 m3 r 7.83

we have L A2 = V1 − V2 = 3.363 × 10−4 m3 − 4.295 × 10−5 m3 = 2.934 × 10−4 m3 where L is the length of the stroke and with d as the bore of the cylinder. The specification calls for

Cycle Efficiency, η

L = 4d

η = 0.500 @ r = 7.83

0.50 0.49 0.48

7.0

7.2

7.4 7.6 Compression Ratio, r

7.8

FIGURE 8.6 Plot of cycle efficiency as a function of the compression ratio for Design Example 2.

8.0

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Introduction to Thermal and Fluid Engineering

Hence, L A2 = (4d)  d=

 4

 d 2 = d 3 = 2.934 × 10−4 m3

2.934 × 10−4 m3 

1/3 = 0.0454 m

(4.54 cm) ⇐

and L = 4d = 4(0.0454 m) = 0.1815 m

8.5

(18.15 cm) ⇐

The Air Standard Diesel Cycle

Two representations of the ideal Diesel cycle are displayed in Figure 8.7. The four strokes in the cycle may be summarized as follows: •

Process 1-2: Isentropic compression

•

Process 2-3: Constant pressure heat addition

•

Process 3-4: Isentropic expansion

•

Process 4-1: Constant volume heat rejection

As in the case of the Otto cycle, we treat all work and heat interactions as positive quantities. For the air standard cycle, the heat transfer during the constant pressure power stroke will be Qin = Q23 = mc p (T3 − T2 ) = m(h 3 − h 2 )

(8.11a)

and the heat rejected during the constant volume exhaust stroke is Qout = Q41 = mc v (T4 − T1 ) = m(u4 − u1 ) P

(8.11b)

T 3 2

3 2

4

4 1 1 s

V (a)

(b)

FIGURE 8.7 The ideal air standard Diesel cycle. (a) the P-V plane and (b) the T-s plane.

Gas Power Systems

233

Consequently, the work for the cycle is Wcycle = Qin − Qout = m(h 3 − h 2 ) − m(u4 − u1 )

(8.11c)

and the cycle thermal efficiency will be

=

Wcycle m(h 3 − h 2 ) − m(u4 − u1 ) u4 − u1 = =1− Qin m(h 3 − h 2 ) h3 − h2

(8.12)

On the cold air standard basis, we take the specific heats and k = c p /c v as constant. The efficiency will be =

mc p (T3 − T2 ) − mc v (T4 − T1 ) T4 − T1 =1− mc p (T3 − T2 ) k(T3 − T2 )

As in the case of the Otto cycle, the compression ratio is the ratio of the cylinder displacement volume to the cylinder clearance volume. In the case of the Diesel cycle, we define the cutoff ratio as the ratio of the cylinder volume after the combustion is completed to the cylinder volume before combustion begins. Hence, the compression and cutoff ratios are defined by r≡

V1 V2

rc ≡

and

V3 V2

The cold air standard efficiency may be put into a form that is a function of only the specific heat ratio, k, and the compression and cutoff ratios, r and rc . In order to accomplish this, we must put all of the temperatures into terms of T1 . First, for the isentropic compression from point 1 to point 2 in Figure 8.7, Equation 8.8a gives T2 = T1



V1 V2

k−1 = r k−1

T2 = T1r k−1

or

and for the constant pressure process between point 2 and point 3 (Figure 8.5) T3 V3 = = rc T2 V2

T3 = T2rc = T1rc r k−1

or

Then, for the isentropic expansion from point 3 to point 4 (Figure 8.7), Equation 8.8b gives T4 = T3



V3 V4

k−1 =

 r k−1 c

r

or

T4 = T1rck

Now, the cold air standard efficiency can be written as  = 1−

T4 − T1 k(T3 − T2 )

= 1−

T1rck − T1 k(T1rc r k−1 − T1r k−1 )

= 1−

rck − 1 k(rc r k−1 − r k−1 )

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Introduction to Thermal and Fluid Engineering

or rk − 1 1  = 1 − k−1c = 1 − k−1 kr (rc − 1) r



rck − 1 k(rc − 1)

 (8.13)

We now illustrate the interplay of the foregoing relationships in a rather lengthy example.

Example 8.3 A Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 takes

in air at an initial pressure of 1 bar and an initial temperature of 27◦ C. For an air standard analysis, determine (a) the pressure and temperature at each if the four points in the cycle, (b) the thermal efficiency, (c) the mean effective pressure, and (d) the cold air standard thermal efficiency.

Solution Assumptions and Specifications (1) The air contained in the piston-cylinder assemblies is an ideal gas and forms a closed system. (2) Kinetic and potential energy changes are negligible. (3) This is an ideal cycle with one constant pressure, one constant volume and two isentropic processes. (4) All of the processes that comprise the cycle are quasi-static. (5) An air standard analysis is specified. (a) At T1 = 27◦ C or 300 K, Table A.12 gives vr 1 = 621.2,

pr 1 = 1.386 and

u1 = 214.07 kJ/kg

Then, to establish conditions at the end of the isentropic compression stroke (point 2 in Figure 8.8), we use the reference volume ratio   vr 2 V2 V2 vr 1 = or vr 2 = vr 1 = vr 1 V1 V1 r P 3

2

4 1 Bar, 27°C

V2 = 16 V1 V3 =2 V2

1

V FIGURE 8.8 P-v diagram for the Example 8.3.

Gas Power Systems

235

so that vr 2 =

621.2 = 38.83 16

and we may use Table A.12 to obtain (with interpolation) T2 = 862.3 K ⇐,

h 2 = 890.95 kJ/kg,

and

pr 2 = 63.77

Then, P2 may be found from  P2 = P1

pr 2 pr 1



 = (1 bar)

63.77 1.386

 = 46.01 bar

or

4.60 MPa ⇐

At the end of the constant pressure heat addition at point 3 in Figure 8.8 P3 = P2 = 4.60 MPa ⇐ and  T3 =

V3 V2

 T2 = rc T2 = 2(862.3 K) = 1724.6 K ⇐

From Table A.12 with interpolation vr 3 = 4.541 and

h 3 = 1910.2 kJ/kg

and for isentropic compression with V4 = V1  vr 4 =

V4 V3



 vr 3 =

V1 V2



V2 V3

 vr 3 =

r vr 3 rc

so that  vr 4 =

 16 (4.541) = 36.32 2

Interpolation of Table A.12 gives T4 = 882.5 K ⇐

and u4 = 660.05 kJ/kg

Finally, because P4 V4 /T4 = P3 V3 /T3  P4 = P3

V3 V4



T4 T3

 = P3

r   T  c 4 r T3

so that  P4 = (4.60 MPa)

2 16



882.5 K 1724.6 K

 = 0.294 bar ⇐

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Introduction to Thermal and Fluid Engineering

(b) The thermal efficiency is given by Equation 8.12  = 1−

u4 − u1 h3 − h2

= 1−

660.05 kJ/kg − 214.07 kJ/kg 1910.2 kJ/kg − 890.95 kJ/kg

= 1−

445.98 kJ/kg 1019.25 kJ/kg

= 1 − 0.438 = 0.562 ⇐ (c) We note that neither the mass of air or the initial volume of the system has been specified and while the mean effective pressure is ordinarily obtained from Equation 8.10, we will need to make a modification that will involve the specific volume. The net work in the cycle is given by Equation 8.11c Wcycle = Qin − Qout = m(h 3 − h 2 ) − m(u4 − u1 )

(8.11c)

and a division by the mass, m gives wcycle = Wcycle /m or wcycle = (h 3 − h 2 ) − (u4 − u1 ) However, h 3 − h 2 and u4 − u1 were obtained in the efficiency computation in part (b) so that wcycle = 1019.25 kJ/kg − 445.98 kJ/kg = 573.27 kJ/kg

Table A.1 gives R = 287 J/kg-K for air so that the inlet specific volume at T1 = 300 K and P1 = 1 bar will be v1 =

RT1 (287 N-m/kg-K)(300 K) = = 0.861 m3 /kg P1 105 N/m2

Now we can put Equation 8.10 on a per kg basis mep =

r Wcycle r wcycle = V1 (r − 1) v1 (r − 1)

so that the mean effective pressure will be mep = =

rwcycle v1 (r − 1) (16)(573.27 kJ/kg) (0.861 m3 /kg)(15)

= = 710.2 kN/m2

or

0.710 MPa ⇐

Gas Power Systems

237

Thermal Efficiency, η

rc = 1 (otto cycle) rc = 2 rc = 3 rc = 4

Compression Ratio, r FIGURE 8.9 The cold air standard thermal efficiency of the Diesel cycle as a function of the compression and cutoff ratios (k = 1.40).

(d) The cold air standard efficiency is given by Equation 8.13  k  1 rc − 1  = 1 − k−1 r k(rc − 1) = 1−

1 160.40



21.40 − 1 (1.40)(2 − 1)



  1 2.639 − 1 = 1− 3.031 1.40 = 1 − 0.386 = 0.614 ⇐ We note that the error involved in the use of the cold air standard amounts to about 9%. A comparison of Equations 8.9 and 8.13 shows that the difference between the thermal efficiency of the Otto and Diesel cycles differs only in the bracketed term that appears in Equation 8.13. Because the bracketed term is always greater than unity, the thermal efficiency of the Otto cycle always exceeds that of the Diesel cycle. As indicated in Figure 8.9, the virtue of the Diesel cycle is in its ability to operate at higher compression ratios. Because V2 = V3 in the Otto cycle, the curve in Figure 8.7 for rc = V3 /V2 = 1 represents the Otto cycle.

8.6

The Gas Turbine

8.6.1 Introduction Figure 8.10a shows the component parts of a gas turbine power system that operates in an open cycle. Work is produced in the turbine, which is on the same shaft as the compressor. Atmospheric air is drawn into the compressor (point 1 in Figure 8.10a) where it is

238

Introduction to Thermal and Fluid Engineering Fuel

Combustion Chamber

Compressor 2

3

Turbine

. Wnet ω 1

4

Air

Products of Combustion (a) . QH Heat Exchanger

Compressor 2

3

Turbine

. Wnet

ω 1

4

. QL

Heat Exchanger

(b) FIGURE 8.10 Component parts of a gas turbine power system (a) the open system and (b) the closed system.

compressed to high pressure and temperature. The air is discharged from the compressor into a combustion chamber (point 2) where fuel is introduced and where combustion occurs to yield combustion products at high temperature. These products of combustion are then led to the turbine (point 3) where they expand to low temperature and pressure and are discharged to the atmosphere at point 4. In order to conduct an air standard analysis of the gas turbine, it is treated as a closed system as indicated in Figure 8.10b. In the air standard analysis, the temperature rise during the combustion process is accounted for by a heat addition from an external source and the return of the working fluid to its initial state is accomplished by a heat rejection to an external sink in the heat exchanger. As in other gas power cycles, the working substance is air, which we take as an ideal gas. 8.6.2 The Ideal Gas Turbine or Brayton Cycle T-s and P-V representations of the ideal air standard Brayton cycle are displayed in Figure 8.11. The cycle contains four processes that may be summarized by: •

Process 1-2: Isentropic compression

•

Process 2-3: Constant pressure heat addition

Gas Power Systems

239

P

T 3 3

4 4

2

2

1

1

s

V (a)

(b)

FIGURE 8.11 The air standard gas turbine or Brayton cycle (a) the P-v and (b) the T-s plane. •

Process 3-4: Isentropic expansion

•

Process 4-1: Constant pressure heat rejection

We note that, because the compressor is on the same shaft as the turbine, in order to obtain the net work of the cycle, the work required to drive the compressor must be subtracted from the work developed by the turbine. For an air standard analysis on an energy rate or power basis, the power developed by the turbine is ˙ turb = W ˙ 34 = m(h W ˙ 3 − h4)

(8.14a)

and the power required to drive the compressor will be ˙ comp = W ˙ 21 = m(h W ˙ 2 − h1)

(8.14b)

Therefore, the power of the cycle is ˙ cycle = W ˙ 34 − W ˙ 21 = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )]

(8.14c)

The heat added in the combustion chamber is ˙ 23 = m(h Q ˙ 3 − h2)

(8.15)

and the thermal efficiency is =

˙ cycle W m[(h ˙ 3 − h 4 ) − (h 2 − h 1 )] = ˙ m(h ˙ 3 − h2) Qin

which can be rearranged to =1−

h4 − h1 h3 − h2

(8.16)

Because a relatively large part of the power developed by the turbine is required to drive the compressor, a useful quantity is the back work ratio (bwr). bwr =

˙ comp W h2 − h1 = ˙ turb h3 − h4 W

(8.17)

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On a cold air standard basis where the specific heats are constant, the efficiency given by Equation 8.16 can be written in terms of the temperatures   T4 − T1 T1 (T4 /T1 ) − 1 =1− =1− T3 − T2 T2 (T3 /T2 ) − 1 and because, as shown in Chapter 7, the temperatures and pressures in an isentropic process are related by T2 = T1



P2 P1

(k−1)/k

T3 = T4

and



P3 P4

(k−1)/k

Moreover, P1 = P2 and P3 = P4 , and we see that T2 T3 = T1 T4

and

T3 T4 = = r (k−1)/k p T2 T1

so that the cold air standard efficiency can be written as =1−

1

(8.18)

(k−1)/k rp

where r p is the pressure ratio that applies to both the turbine and compressor. rp ≡

P2 P3 ≡ P1 P4

(8.19)

Example 8.4 An ideal Brayton cycle takes in air at a volumetric flow rate of 300,000 L/min

at 1 bar and 27◦ C. The pressure ratio in the isentropic compression and expansion is 6 and the products of combustion leave the combustion chamber at 927◦ C (Figure 8.12). For the air standard analysis, determine (a) the net power developed by the turbine, (b) the back work ratio, and (c) the thermal efficiency. Then determine (d) the cold air standard thermal efficiency.

T

927°C

2

1

3

4

1 Bar, 27°C, 300,000 L/min P2 =6 P1 s

FIGURE 8.12 T-s diagram for Example 8.4.

Gas Power Systems

241

Solution Assumptions and Specifications (1) The air is an ideal gas and the system is an open system. (2) Kinetic and potential energy changes throughout the cycle are negligible. (3) The cycle consists of two isentropic and two constant pressure processes. (4) An air standard analysis is required. We determine the required temperatures and enthalpies from Table A.12 using Figure 8.10. At T1 = 27◦ C (300 K), we read pr 1 = 1.386

and

h 1 = 300.19 kJ/kg

and then because pr 2 P2 = = rp = 6 pr 1 P1 we see immediately that the reference pressure at the end of the isentropic compression (point 2) is pr 2 = pr 1r p = (1.386)(6) = 8.316 Then to establish the conditions at the end of the isentropic compression, we use Table A.12 (with interpolation) T2 = 498.4 K and

h 2 = 501.27 kJ/kg

The temperature of the products of combustion at the end of the constant pressure heat addition (point 3) is specified as 1200 K. Table A.12 gives pr 3 = 238.0 and

h 3 = 1277.79 kJ/kg

Then because pr 4 P4 1 1 = = = pr 3 P3 rp 6 we find that the reference pressure at point 4 will be pr 4 =

pr 3 238.0 = = 39.67 rp 6

With interpolation, Table A.12 provides T4 = 761.9 K and

h 4 = 780.32 kJ/kg

˙ ˙ is the volumetric (a) Table A.1 gives us R = 287 J/kg-K and because m ˙ = V/v where V flow rate   1 ˙ = (300,000 L/min)(10−3 m3 /L) V = 5 m3 /s 60 s/min

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Hence, using conditions at point 1 gives m ˙ = =

˙ P1 V RT1 (5 m3 /s)(1 × 105 N/m2 ) (287 N-m/kg-K)(300 K)

= 5.807 kg/s On an air standard basis, the net power developed by the turbine is found from Equation 8.14a ˙ turb = m(h W ˙ 3 − h4) = (5.807 kg/s(1277.79 kJ/kg − 780.32 kJ/kg) = (5.807 kg/s)(497.47 kJ/kg) = 2888.8 kW ⇐ (b) The back work ratio given by Equation 8.17 requires a computation of the work chargeable to the compressor given by Equation 8.14b ˙ comp = m(h W ˙ 2 − h1) = (5.807 kg/s)(501.27 kJ/kg − 300.19 kJ/kg) = (5.807 kg/s)(201.08 kJ/kg) = 1167.67 kW The back work ratio is bwr =

˙ comp W 1167.67 kW = = 0.404 ⇐ ˙ 2888.8 kW Wturb

(c) The evaluation of the thermal efficiency via Equation 8.16 requires that the net work of the cycle ˙ cycle = W ˙ turb − W ˙ comp W = 2888.8 kW − 1167.67 kW = 1721.1 kW be divided by the heat input given by Equation 8.15 ˙ in = m(h Q ˙ 3 − h2) = (5.807 kg/s)(1277.79 kJ/kg − 501.27 kJ/kg) = (5.807 kg/s)(776.52 kJ/kg) = 4509.3 kW Hence, =

˙ cycle W 1721.1 kW = 0.382 = ˙ in 4509.3 kW Q

or

38.2% ⇐

Gas Power Systems

243 y

s . Qin

Compressor

Combustion Chamber

x

2

4

Turbine

3 . W

ω

1 (a) 3

T

x 4

2 y

1

s

(b)

FIGURE 8.13 The ideal gas turbine or Brayton cycle with regeneration (a) the system and (b) the T-s diagram.

(d) The cold air standard efficiency is given by Equation 8.18. With k = 1.40 =1−

1 (k−1)/k rp

=1−

1 60.40/1.40

=1−

1 = 1 − 0.599 = 0.401 ⇐ 1.669

We note in this particular example, that the error involved in using the cold air standard efficiency is about 5%. 8.6.3 The Ideal Brayton Cycle with Regeneration In the Brayton cycle, the products of combustion are exhausted from the turbine at a relatively high temperature. It is possible to reduce the consumption of fuel in the combustion chamber by preheating the air leaving the compressor with energy obtained from the turbine exhaust. This process takes place in a heat exchanger, known as a recuperator, and a system with regeneration is shown in Figure 8.13a.3 We see that the working fluid enters the cold side of the recuperator at point 2 and passes through the recuperator to enter the 3 The

process of regeneration applies to the heating of the air leaving the compressor. This heating is done in a heat exchanger called a recuperator.

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Introduction to Thermal and Fluid Engineering

combustion chamber at point x. The spent air-fuel mixture, which still leaves the turbine at point 4, passes through the hot side of the recuperator, and leaves the recuperator at point y. The T-s diagram for this cycle is shown in Figure 8.13b. The recuperator itself is an internally reversible heat exchanger where the fluid, in passing through it, experiences no pressure loss due to friction. We may note in Figure 8.13 that the maximum possible heat transfer in the recuperator would occur if gas entering the combustion chamber were at the exhaust temperature from the turbine: ˙ max = m(h Q ˙ 4 − h2)

(8.20a)

In actual practice, however, because achievement of the turbine exhaust temperature would require a recuperator of infinite size, point x is at a temperature that is lower than the temperature at point 4. The actual heat transferred in the recuperator is therefore, ˙ act = m(h Q ˙ x − h2)

(8.20b)

Equations 8.20 can be employed to define a recuperator effectiveness =

˙ act Q m(h ˙ x − h2) h x − h2 = = ˙ max m(h ˙ 4 − h2) h4 − h2 Q

(8.21a)

and if the recuperator effectiveness is specified, we can determine the enthalpy at point x via h x = h 2 + (h 4 − h 2 )

(8.21b)

An effectiveness value of 1.00 (or 100%) means that the temperature of the air stream entering the combustion chamber is equal to the temperature of the combustion products leaving the turbine. In practice, however, recuperator effectiveness values range from 0.60 to 0.80 (60% to 80%) because, as already stated, higher effectiveness values are only attainable in rather impractical large sizes. The net power developed by the cycle remains unchanged when the recuperator is employed and is still given by Equation 8.14c ˙ cycle = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )]

(8.14c)

However, when regeneration is employed, the heat input is ˙ in = m(h Q ˙ 3 − hx)

(8.22)

and the thermal efficiency of the cycle is =

˙ cycle W (h 3 − h 4 ) − (h 2 − h 1 ) = ˙ h3 − h x Qin

(8.23)

It can be shown (Problem 8.39 asks for a derivation) that when the compressor and turbine inlet temperatures are fixed at T1 and T3 , the net work is a maximum when T2 = (T1 T3 ) 1/2

(8.24)

and the pressure ratio for this condition is given by P2 = P1



T2 T1

k/2(k−1) (8.25)

We now use Example 8.5 to show the improvement in thermal efficiency when a recuperator is employed.

Gas Power Systems

245 T

3

x 4 2

y

1 Bar, 27°C, 300,000 L/min 1 P2 = 6, ε = 0.78 P1 s FIGURE 8.14 T-s diagram for Example 8.5.

Example 8.5 Once again, consider the ideal Brayton cycle of Example 8.4 and suppose

that a recuperator with an effectiveness of  = 0.78 is placed in the system as shown in Figure 8.14. For an air standard analysis determine the thermal efficiency and compare it with the value obtained in Example 8.4.

Solution The same assumptions made in Example 8.4 will be made here. The enthalpy values at the four points in the cycle that were found in Example 8.4 are summarized as h 1 = 300.19 kJ/kg

h 2 = 501.27 kJ/kg

h 3 = 1277.79 kJ/kg

h 4 = 780.32 kJ/kg

Point x is at the exit of the cold air-side of the recuperator and its enthalpy can be found from Equation 8.21b h x = h 2 + (h 4 − h 2 ) = 501.27 kJ/kg + (0.78)(780.32 kJ/kg − 501.27 kJ/kg) = 501.27 kJ/kg + (0.78)(279.05 kJ/kg) = 501.27 kJ/kg + 217.66 kJ/kg = 718.93 kJ/kg The net power developed by the cycle is the same as the value determined in Example 8.4 ˙ cycle = 1721.1 kW W Here, however, the heat input will be ˙ in = m(h Q ˙ 3 − hx) = (5.807 kg/s)(1277.79 kJ/kg − 718.93 kJ/kg) = (5.807 kg/s)(558.86 kJ/kg) = 3245.3 kW

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Introduction to Thermal and Fluid Engineering

and the thermal efficiency will be =

˙ cycle W 1721.1 kW = = 0.530 ˙ in 3245.31 kW Q

or

53.0% ⇐

The marked improvement in the thermal efficiency from 0.382 to 0.530 may be noted.

8.7

The Jet Engine

Because of its excellent power-to-weight characteristics, the gas turbine has been adapted for use in aircraft propulsion. As indicated in the cutaway view of the turbojet engine of Figure 8.15a, the power developed by the turbine is just sufficient to drive the compressor and other auxiliary components. The components of a turbojet engine are indicated in Figure 8.15a and the T-s diagram for the ideal turbojet engine is shown in Figure 8.15b. The engine itself is composed of three main sections: the compressor, the combustion chamber or combustor, and the turbine. Air, treated as an ideal gas, is the working medium and enters the diffuser section (located just upstream of the compressor) at atmospheric conditions. The purpose of the diffuser is to Combustion Chamber

Compressor

3

2

1

Turbine

4

5

Diffuser

5

Nozzle (a)

4 T P=C

5 6

3 2 P=C

1

(b) FIGURE 8.15 A turbojet engine (a) the cycle and (b) the T-s plane.

s

Gas Power Systems

247

decelerate the air entering the engine and energy absorbed during this deceleration leads to a pressure rise, which is referred to as the ram effect. The compressor, combustor, and turbine serve the same function as they do in the gas turbine. The air-fuel mixture leaving the turbine at a pressure that is significantly higher than atmospheric is led to the nozzle where an expansion to low pressure and high velocity takes place. It is this change in velocity between point 1 and point 6 (see Figure 8.15b) that gives rise to the thrust of the engine. There are six points on the T-s diagram of Figure 8.18b that represent the ideal cycle and define the six processes that comprise the cycle. These processes are described by •

Process 1-2: This is an isentropic compression in the diffuser where the velocity of the incoming air is reduced.

•

Process 2-3: This is also an isentropic compression. It occurs in the compressor where the pressure is increased through the pressure ratio, P3 P2

rp = •

Process 3-4: This is a constant pressure heat addition in the combustor or burner section. Here fuel is injected and the products of combustion leaving at point 4 are at high pressure and temperature.

•

Process 4-5: This is an isentropic expansion in the turbine where the pressure and temperature of the products of combustion are reduced. However, we have noted that the pressure at the exhaust from the turbine is significantly greater than atmospheric and this permits the nozzle to provide a high velocity at its exhaust.

•

Process 5-6: This is an isentropic process in the nozzle where the products of combustion leave at high velocity and are discharged to the surroundings.

•

Process 6-1: This is a constant pressure heat rejection that returns the cycle to its starting point.

We begin the analysis of the turbojet engine by noting that the thrust developed by the engine is ˆ6−V ˆ 1) FT = m( ˙ V

(N)

(8.26)

where m ˙ is the air flow rate. Here, the velocity leaving the nozzle may be developed from a first law energy balance between points 5 and 6 in Figure 8.15. In the absence of work done on or by the air, heat transferred, potential energy changes, and neglible velocity of the air at point 5, the energy rate balance will be h5 = h6 +

Vˆ 26 2

so that Vˆ 6 = [2(h 5 − h 6 )]1/2

( m/s)

(8.27)

The propulsive power developed by the engine is the product of this thrust and the velocity of the aircraft ˆ ac = m( ˆ6−V ˆ 1) V ˆ ac ˙ P = FT V W ˙ V

(W)

(8.28)

The heat supplied in the combustor is ˙ in = m(h Q ˙ 4 − h3)

(W)

(8.29)

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Introduction to Thermal and Fluid Engineering

and, because the net work developed by the engine is zero, the propulsion efficiency is defined as the ratio of the propulsive power to the energy released by the fuel P =

ˆ6−V ˆ 1) V ˆ ac ˙P W (V = ˙ in h4 − h3 Q

(8.30)

We now turn to a rather comprehensive example where we resort to the first law energy equation for a control volume to help us generate the solution.

Example 8.6 A jet aircraft is flying at a velocity of 1000 km/h at an altitude where the air

temperature is −40◦ C and the pressure is 0.40 bar. The compressor has a pressure ratio of 10 and the temperature at the entrance to the turbine is 727◦ C. The inlet to the engine has an area of 0.579 m3 . Using an air standard analysis and an ideal Brayton cycle, determine (a) the temperature of the air leaving the diffuser, (b) the pressure of the air leaving the compressor, (c) the temperature of the air leaving the nozzle, (d) the velocity of the air leaving the nozzle, (e) the thrust produced by the engine, and (f) the propulsion efficiency of the engine.

Solution Assumptions and Specifications (1) The air is an ideal gas and forms a closed system. (2) The cycle is an ideal cycle consisting of two isentropic processes and two constant pressure processes. (3) Each component of the system can be analyzed as a control volume at steady state. (4) Potential energy changes are negligible. (5) The diffuser and the compressor undergo isentropic compressions and the turbine and the nozzle undergo isentropic expansions. (6) All processes are quasi-static. (7) The turbine produces just enough power to operate the compressor. (8) An air standard analysis has been specified. (a) For the diffuser, the inlet pressure is P1 = 0.40 bar and the inlet temperature is 233 K (−40◦ C). Table A.12 (with interpolation) gives pr 1 = 0.5740

and

h 1 = 233.02 kJ/kg

The evaluation of the temperature leaving the diffuser (point 2 in Figure 8.16) requires consideration of the diffuser as a control volume. An energy rate balance with no heat or work interactions across the control volume boundaries and no change in either kinetic or potential energies and with negligible exit velocity produces ˆ2 V h2 = h1 + 1 2 and with Vˆ1 =

106 m/h = 277.78 m/s 3600 s/h

Gas Power Systems

249 T4 = 727°C 4 T

5 6

3 P3 = 10 P2

2 1

T1 = –40°C P1 = 0.40 bar V1 = 1000 km/m A1 = 0.579 m2 s

FIGURE 8.16 T-s diagram for Example 8.6.

the value of h 2 will be h2 = h1 +

2 Vˆ1 2

= 233.02 kJ/kg +

(277.78 m/s) 2 (1 N-s2 /kg-m) 2(1000 J/kJ)

= 233.02 J/kg + 38.58 kJ/kg = 271.60 kJ/kg We can then interpolate Table A.12 at h 2 = 271.60 kJ/kg to obtain T2 = 271.5 K ⇐ and then

 P2 = P1

pr 2 pr 1

and



 = (0.40 bar)

pr 2 = 0.9783

0.9783 0.5740

 = 0.6818 bar

We are also able to obtain the mass flow of the air and select R from Table A.1 as 287 J/kgK. Then using conditions at point 1, we have   P1 ˆ ˆ1 m ˙ = 1 A1 V 1 = A1 V RT1 Thus,

 m ˙ =

4 × 104 N/m2 (0.579 m2 )(277.78 m/s) = 96.21 kg/s (287J/kg-K)(233 K)

(b) In seeking the compressor exit temperature at point 3 in Figure 8.16, we begin by noting that P3 /P2 = 10 so that P3 = 10P2 = 10(0.6818 bar) = 6.818 bar ⇐

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Introduction to Thermal and Fluid Engineering

and

 pr 3 = pr 2

P3 P2

 = (0.9783)(10) = 9.783

The temperature and enthalpy at point 3 are obtained from Table A.12 with interpolation at pr 3 = 9.783 T3 = 521.4 K

and

h 3 = 525.12 K

(c) At point 4 (the inlet to the turbine), the temperature is specified as 1000 K (727◦ C) and from Table A.12 pr 4 = 114.0

and

h 4 = 1046.04 kJ/kg

and because the power developed by the turbine is just sufficient to run the compressor, we have ˙ turb = W ˙ comp W m(h ˙ 4 − h 5 ) = m(h ˙ 3 − h2) Solving for h 5 h 5 = h 4 − (h 3 − h 2 ) = 1046.04 kJ/kg − (525.12 kJ/kg − 271.60 kJ/kg) = 1046.04 kJ/kg − 253.52 kJ/kg = 792.51 kJ/kg Interpolation in Table A.12 then yields the values of T5 and pr 5 T5 = 773.1 kJ/kg ⇐

and

pr 5 = 41.95

and because P3 = P4 = 6.818 bar     pr 5 41.95 P5 = P4 = (6.818 bar) = 2.509 bar pr 4 114.0 The expansion through the nozzle is isentropic to atmospheric pressure. Thus, at point 6     P6 0.40 bar pr 6 = pr 5 = (41.95) = 6.687 P5 2.509 bar and this time, Table A.12 gives, with interpolation T6 = 468.9 K ⇐

and

h 6 = 471.11 kJ/kg

(d) With just the nozzle as a control volume, the energy rate balance between points 5 and 6 in Figure 8.16 can be written in the absence of heat and work interactions across the boundaries of the control volume, the absence of potential energy change and negligible inlet velocity

 ˆ2 V 6 m ˙ h6 + = mh ˙ 5 2 and the exit velocity is seen to be ˆ 6 = [2(h 5 − h 6 )]1/2 V

Gas Power Systems

251

and with h 5 − h 6 = 792.51 kJ/kg − 471.11 kJ/kg = 321.40 kJ/kg then ˆ 6 = [2(321, 400 N-m/kg)(1 kg-m/N-s2 )]1/2 V = [642.80 m2 /s2 ]1/2 = 801.75 m/s ⇐ (e) The thrust is given by Equation 8.26 ˆ6−V ˆ 1) FT = m( ˙ V = (96.21 kg/s)(801.75 m/s − 277.78 m/s) = (96.21 kg/s)(523.97 m/s) = 50,410 N ⇐ (f) The propulsive power is given by Equation 8.28 ˆ6−V ˆ 1) V ˆ ac = FT V ˆ ac ˙ P = m( W ˙ V = (50, 410 N)(277.78 m/s) = 14.00 MW and the heat added in the combustor is given by Equation 8.29 ˙ in = m(h Q ˙ 4 − h3) = (96.21 kg/s)(1046.04 kJ/kg − 525.12 kJ/kg) = (96.21 kg/s)(520.92 kJ/kg) = 5.012 × 104 kW Then, the propulsion efficiency, given by Equation 8.29, will be P =

8.8

˙ cycle W 1.400 × 107 W = = 0.279 ˙ in 5.011 × 107 W Q

or

27.9% ⇐

Summary

Air standard analysis of gas power systems involves air as the working medium where both the ideal gas law and variable specific heats are employed. Internal energy and enthalpy values are obtained from the “air tables” (given here as Table A.12) and are based on the use of the reference volume and reference pressure vri Vi = vr j Vj

and

pri Pi = pr j Pj

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Introduction to Thermal and Fluid Engineering

Cold air standard analysis of gas power systems involves air as the working medium where the ideal gas law and constant specific heats are employed. In this case, the specific heats are taken from Table A.1. In the ideal air standard Otto or gasoline engine cycle, there are four processes that are shown in Figure 8.2. The compression ratio is r=

V1 V4 = V2 V3

the heat added is Qin = m(u3 − u2 )

(8.3a)

Qout = m(u4 − u1 )

(8.3b)

Wcycle = m[(u3 − u4 ) − (u2 − u1 )]

(8.4a)

the heat rejected is

the net work of the cycle is

and the efficiency is =1−

u4 − u1 u3 − u2

(8.5)

The cold air standard efficiency is =1−

1 r k−1

(8.9)

In the ideal air standard diesel engine cycle, there are four processes as shown in Figure 8.7. The heat input is given by Qin = Q23 = mc p (T3 − T2 ) = m(h 3 − h 2 )

(8.11a)

Qout = Q41 = mc v (T4 − T1 ) = m(u4 − u1 )

(8.11b)

Wcycle = Qin − Qout = m(h 3 − h 2 ) − m(u4 − u1 )

(8.11c)

and the heat rejected is

The work done is

and the thermal efficiency will be =1−

u4 − u1 h3 − h2

On the cold air standard basis, the efficiency will be   1 rck − 1  = 1 − k−1 r k(rc − 1)

(8.12)

(8.13)

In the ideal air standard Brayton or gas turbine cycle, there are four processes as shown in Figure 8.10a.

Gas Power Systems

253

The heat added in the combustion chamber is ˙ in = m(h Q ˙ 3 − h2)

(8.15)

˙ cycle = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )]

(8.14c)

The power of the cycle is

and the thermal efficiency is =1−

h4 − h1 h3 − h2

(8.16)

bwr =

h2 − h1 h3 − h4

(8.17)

The back work ratio is

The cold air standard efficiency is =1−

1

(8.18)

(k−1)/k rp

where r p is the pressure ratio rp =

P2 P3 = P1 P4

(8.19)

In the ideal air standard Brayton with regeneration, the cycle contains a recuperator as shown in Figure 8.13, when the recuperator effectiveness is specified, the enthalpy at point x can be determined from h x = h 2 + (h 4 − h 2 )

(8.21b)

The net power developed by the cycle is given by ˙ cycle = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )]

(8.14c)

˙ in = m(h Q ˙ 3 − hx)

(8.22)

the heat input is

and the thermal efficiency of the cycle is =

(h 3 − h 4 ) − (h 2 − h 1 ) h3 − h x

(8.23)

In the ideal air standard jet engine cycle shown, there are six processes as shown in Figure 8.15. The thrust developed by the engine is ˆ6−V ˆ 1) FT = m( ˙ V

(N)

(8.26)

The velocity at the exit from the nozzle will be Vˆ 6 = [2(h 5 − h 6 )]1/2

(8.27)

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Introduction to Thermal and Fluid Engineering

The propulsive power developed by the engine is ˆ ac = m( ˆ6−V ˆ 1) V ˆ ac ˙ P = FT V W ˙ V

(8.28)

The heat supplied in the combustor will be Qin = m(h ˙ 4 − h3)

(8.29)

and the propulsion efficiency is P =

8.9

ˆ6−V ˆ 1) V ˆ ac (V h4 − h3

(8.30)

Problems

Use of the Air Tables 8.1: Air at 37◦ C and 120 kPa is compressed isentropically to a pressure of 1.50 MPa. Determine the final temperature and enthalpy for (a) an air standard process and (b) a cold air standard process. 8.2: Air at 42◦ C and 125 kPa is compressed isentropically through V1 /V2 = 8. Determine the final temperature and pressure for (a) an air standard process and (b) a cold air standard process. 8.3: In a constant volume process, 600 kJ of heat is added to air at 700 K and 16 bar. Determine the final temperature and pressure for (a) an air standard process and (b) a cold air standard process. 8.4: In a constant pressure process, 650 kJ of heat is added to air at 600 K and a volume of 0.40 m3 . Determine the final temperature and volume for (a) an air standard process and (b) a cold air standard process. 8.5: An air standard cycle (specific heats are variable) is composed of four processes. Process 1-2 is an isentropic compression from P1 = 100 kPa and T1 = 300 K to P2 = 850 kPa, process 2-3 is a constant volume heat addition to T3 = 1750 K, process 34 is an isentropic expansion to 100 kPa, and process 4-1 is a constant pressure heat rejection. Determine (a) the heat added, (b) the heat rejected, (c) the net work, and (d) the thermal efficiency. 8.6: Rework Problem 8.5 using the cold air standard (constant specific heats taken at 300 K). 8.7: An air standard cycle (specific heats are variable) is composed of three processes. Process 1-2 is a constant volume heat addition to T1 = 22◦ C and 98 kPa to 392 kPa, process 2-3 is an isentropic expansion to 98 kPa and process 3-1 is a constant pressure heat rejection to the initial state. Assume that 0.005 kg of air is present and determine (a) the heat added, (b) the heat rejected, (c) the net work, and (d) the thermal efficiency. 8.8: Rework Problem 8.7 using a cold air standard cycle (constant specific heats taken at 300 K). The Ideal Otto Cycle 8.9: An air standard Otto cycle has a compression ratio of 8.75 with inlet conditions of 100 kPa and 27◦ C and the heat added is 1325 kJ/kg. For a mass of 1 kg, determine (a) the heat rejection, (b) the net work, (c) the thermal efficiency, and (d) the mean effective pressure.

Gas Power Systems

255

8.10: Rework Problem 8.9 using a cold air standard cycle (constant specific heats taken at 300 K). 8.11: Intake conditions of an air standard Otto cycle are P1 = 100 kPa, V1 = 420 cm3 , and T1 = 22◦ C. If the maximum temperature in the cycle is to be limited to 2250 K and the compression ratio is 8, determine (a) the mass of air present, (b) the heat added, (c) the heat rejected, (d) the net work, (e) the thermal efficiency, and (f) the mean effective pressure. 8.12: Rework Problem 8.11 using a cold air standard cycle (constant specific heats taken at 300 K) to find the temperature at each point in the cycle and the thermal efficiency. 8.13: An air standard Otto cycle has a temperature of 500 K and a pressure of 600 kPa at the end of the isentropic expansion. For a compression ratio of 7.5 and 160 kJ/kg of heat rejected from the cycle, determine (a) the heat input, (b) the net work done, and (c) the thermal efficiency. 8.14: Rework Problem 8.13 using a cold air standard cycle (constant specific heats taken at 300 K) to find the temperature at each point in the cycle and the thermal efficiency. 8.15: An ideal Otto cycle has a compression ratio of 8 and the inlet conditions are T1 = 32◦ C and P1 = 98 kPa. The cycle heat addition is 725 kJ/kg. Use an air standard analysis to determine (a) the maximum pressure in the cycle, (b) the heat rejected, (c) the net work, (d) the thermal efficiency, and (e) the mean effective pressure. 8.16: Rework Problem 8.15 using a cold air standard analysis (constant specific heats taken at 300 K). 8.17: In an Otto cycle where the temperature at the end of the isentropic expansion is 820 K, the compression ratio is 8 and the inlet conditions are P1 = 100 kPa, V1 = 625 cm3 , and T1 = 27◦ C. Use an air standard analysis to determine (a) the highest temperature and pressure in the cycle, (b) the heat added, (c) the heat rejected, (d) the net work, (e) the thermal efficiency, and (f) the mean effective pressure. 8.18: Use the conditions of Problem 8.17 and use a cold air standard analysis (constant specific heats taken at 300 K) to determine the maximum temperature and pressure in the cycle. 8.19: Consider an ideal Otto cycle operating on the air standard with a compression ratio of 8.6. The minimum and maximum temperatures that occur in the cycle are 37◦ C and 827◦ C. The pressure at the inlet is 100 kPa and 0.640 kJ are added to the charge during each cycle. Determine (a) the heat rejected, (b) the net work done, and (c) the thermal efficiency. 8.20: Rework Problem 8.19 using a cold air standard cycle (constant specific heats taken at 300 K). 8.21: In an air standard Otto cycle with a compression ratio of 8.75, the pressure during the constant volume heat addition is tripled. If the inlet conditions are P1 = 100 kPa and T1 = 27◦ C, determine (a) the temperature at points 1 and 4 shown in Figure 8.2 and (b) the thermal efficiency. 8.22: Rework Problem 8.21 using a cold air standard cycle (constant specific heats taken at 300 K). The Ideal Diesel Cycle 8.23: The pressure and temperature at the beginning of the isentropic compression in an air standard Diesel cycle are 96 kPa and 17◦ C. At the end of compression, the pressure is 6.45 MPa and the temperature is 2000 K. Determine (a) the compression ratio, (b) the cutoff ratio, (c) the thermal efficiency, and (d) the mean effective pressure.

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Introduction to Thermal and Fluid Engineering

8.24: Rework Problem 8.23 using a cold air standard cycle (constant specific heats taken at 300 K). 8.25: An air standard Diesel cycle has a compression ratio of 16. Determine the thermal efficiencies for cutoff ratios of 1 (Otto cycle), 2, 3, 4, and 5. 8.26: The pressure and temperature at the beginning of the isentropic compression in an air standard Diesel cycle are 185 kPa and 102◦ C. The compression ratio is 20 and the heat addition is 908 kJ/kg. Determine (a) the maximum temperature and pressure in the cycle, (b) the cutoff ratio, (c) the net work in kJ/kg, and (d) the thermal efficiency. 8.27: Rework Problem 8.26 using a cold air standard cycle (constant specific heats taken at 300 K). 8.28: An air standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. Conditions at the beginning of the isentropic compression are P1 = 96 kPa, T1 = 287 K, and V1 = 0.045 m3 . Determine (a) the heat added in kJ, (b) the heat rejected in kJ, (c) the thermal efficiency, and (d) the mean effective pressure. 8.29: Rework Problem 8.28 using a cold air standard cycle (constant specific heats taken at 300 K). 8.30: A Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.5 draws air at 100 kPa and 300 K. Using a cold air standard analysis, determine (a) the pressures and temperatures at states 2, 3, and 4 in Figure 8.7, (b) the thermal efficiency, and (c) the mean effective pressure. 8.31: Rework Problem 8.30 using an air standard analysis. 8.32: A Diesel engine is to be designed to produce a cold air standard efficiency of 55% when operating with a compression ratio of 15. Determine the cutoff ratio for the engine. 8.33: Use L’Hospital’s rule to show that the efficiency of a cold air standard Diesel cycle as given by Equation 8.13 reduces to that of the Otto cycle in the limit when rc −→ 1. 8.34: In an ideal air standard Diesel cycle, the compression ratio is 15 and the heat addition per cycle is 10.8 kJ. At the beginning of the isentropic compression, P1 = 97.5 kPa, T1 = 20◦ C, and V1 = 0.015 m3 . Determine (a) the mass of air, (b) the maximum temperature in the cycle, (c) the net work, and (d) the thermal efficiency. 8.35: Rework Problem 8.34 using a cold air standard cycle. 8.36: A cold air standard Diesel cycle with a compression ratio of 15 has a maximum temperature of 1600 K and a minimum temperature of 300 K. The thermal efficiency of the cycle is 80% of that of a Carnot engine operating between the same temperature limits. Determine the cutoff ratio. The Ideal Gas Turbine Cycle without Regeneration 8.37: A gas turbine operating on a Brayton cycle takes in air at 290 K and operates with a pressure ratio of 8. The mass flow through the turbine is 35 kg/s and the turbine inlet temperature is 1250 K. Assume operation on the air standard and determine (a) the power required to drive the compressor, (b) the back work ratio, and (c) the thermal efficiency. 8.38: Rework Problem 8.37 using the cold air standard. 8.39: Show that the net work is at a maximum in the Brayton cycle with fixed compressor and turbine inlet temperatures (T1 and T3 ) when T2 = (T1 T3 ) 1/2

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257

and the pressure ratio for this condition is given by P2 = P1



T3 T1

k/2(k−1)

8.40: The minimum and maximum temperatures in a Brayton cycle are 300 K and 1400 K, respectively. Using the cold air standard, determine (a) the pressure ratio that produces the maximum net work, (b) the compressor work, (c) the turbine work, (d) the heat supplied in the combustion chamber, and (e) the thermal efficiency. 8.41: A gas turbine operates on a Brayton cycle with a pressure ratio of 7 for both the compressor and the turbine. The compressor inlet conditions are 100 kPa and 300 K and the turbine inlet temperature is 1500 K. The isentropic efficiencies of the compressor and the turbine are 0.85 and 0.95, respectively. Using the cold air standard and a mass flow of 25 kg/s, determine (a) the compressor power input, (b) the rate of heat supply in the combustion chamber, (c) the turbine output power, and (d) the thermal efficiency. 8.42: Rework Problem 8.41 using the air standard. 8.43: The compressor pressure ratio of an ideal air standard Brayton cycle is 10. If air enters the compressor at 101.3 kPa and 295 K with a mass flow rate of 11.50 kg/s, determine (a) the net power developed and (b) the thermal efficiency. 8.44: Rework Problem 8.43 using the cold air standard. 8.45: The compressor pressure ratio of an ideal air standard Brayton cycle is 12.5 and the net power developed is 7350 kW. If the maximum and minimum temperatures in the cycle are 1560 K and 290 K, respectively, determine (a) the mass flow rate of the air and (b) the thermal efficiency. 8.46: Rework Problem 8.45 using the cold air standard. 8.47: In an ideal air standard Brayton cycle, air enters the compressor at 101.3 kPa and 25◦ C with a volumetric flow rate of 4.8 m3 /s. The turbine inlet is at 1425 K and the compressor pressure ratio is 7.5. Determine (a) the thermal efficiency, (b) the back work ratio, and (c) the net power developed in kW. 8.48: Rework Problem 8.47 with a compressor pressure ratio of 10. 8.49: Rework Problem 8.47 with a compressor pressure ratio of 15. 8.50: Rework Problem 8.47 with a turbine inlet temperature of 1550 K. 8.51: Rework Problem 8.47 with a turbine inlet temperature of 1600 K. 8.52: Rework Problem 8.43 with compressor and turbine isentropic efficiencies of 0.85. The Ideal Gas Turbine Cycle with Regeneration 8.53: Suppose that a recuperator with an effectiveness of 78% is incorporated into the Brayton cycle of Problem 8.37. Determine (a) the net power developed in kW, (b) the back work ratio, and (c) the thermal efficiency. 8.54: Rework Problem 8.55 with a recuperator effectiveness of 0.84. 8.55: Rework Problem 8.54 with a recuperator effectiveness of 0.90. 8.56: Show that for an ideal Brayton cycle with a one hundred percent effectiveness recuperator, the thermal efficiency is given by =1−

T1 (r p ) (k−1)/k T3

258

Introduction to Thermal and Fluid Engineering where r p is the pressure ratio, T1 is the compressor inlet temperature, and T3 is the turbine inlet temperature.

8.57: Data for a gas turbine engine operating on the Brayton cycle equipped with a recuperator is Pressure and temperature at the compressor inlet: 100 kPa and 298 K Pressure ratio: r p = 10 Mass flow rate: 10 kg/s Turbine inlet temperature: 1600 K Recuperator effectiveness:  = 0.70 Using the cold air standard, determine (a) the rate of heat supply in the combustion chamber and (b) the thermal efficiency of the engine. 8.58: Rework Problem 8.53 with an isentropic compressor efficiency of 0.90 with turbine operation remaining isentropic. 8.59: Rework Problem 8.53 with an isentropic turbine efficiency of 0.90 with compressor operation remaining isentropic. 8.60: Rework Problem 8.53 with isentropic compressor and turbine efficiencies of 0.90. 8.61: A gas turbine with a recuperator operates in an air standard Brayton cycle. Air enters the compressor at 96.5 bar and 300 K and is compressed to 482.4 bar. After the air passes through the regenerator its temperature is 600 K. Both the compressor and turbine each have isentropic efficiencies of 0.875 and the air enters the turbine at a mass flow rate of 35 kg/s. Determine (a) the thermal efficiency of the cycle, (b) the recuperator effectiveness, and (c) the volumetric flow rate of the air entering the compressor in m3/s. 8.62: Rework Problem 8.61 for an air temperature at the outlet of the regenerator of 640 K. The Jet Engine 8.63: A turbojet flying at 330 m/s and 238 K draws air at 40 kPa into its diffuser section. The air leaving the diffuser section enters a compressor that provides a pressure ratio of 6.5. The temperature of the air entering the turbine is 1200 K and the mass flow rate through the engine is 70 kg/s. Assuming an ideal jet engine cycle as described in Section 8.7 and using the cold air standard, determine (a) the temperature and pressure of the air leaving the diffuser, (b) the temperature and pressure of the air at the compressor outlet, (c) the rate of heat supply in the combustion chamber, (d) the temperature and velocity of the air at the nozzle exit, (e) the thrust developed by the engine, and (f) the propulsion efficiency of the engine. 8.64: A turbojet flying at 350 m/s draws air at 50 kPa and 237 K into its diffuser section. The air leaves the diffuser with negligible velocity and enters a compressor that provides a pressure ratio of 8.75. The temperature of the air entering the turbine is 1200 K and the mass flow rate through the engine is 72 kg/s. Assuming an ideal jet engine as described in Section 8.7 and using the cold air standard, determine (a) the temperature and pressure of the air leaving the diffuser, (b) the temperature and pressure of the air at the compressor inlet, and (c) the rate of heat supply in the combustion chamber. 8.65: Consider the ideal jet engine without the diffuser (that is, with no ram effect). The compressor draws 50 kg/s of air at 75 kPa and 280 K and compresses it to 525 kPa. The heat suppled to the combustion chamber, which burns a fuel having a heating value of 41,800 kJ/kg, raises the temperature of the air to 1500 K. Assume the cold air

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259

standard and determine (a) the temperature and pressure of the air at the exit from the turbine, (b) the mass flow rate of the fuel required, and (c) the velocity at exit from the nozzle if the exit pressure is 70 kPa. 8.66: Rework Problem 8.65 using the air standard. 8.67: Air at 20 kPa and −23◦ C enters a turbojet engine with a velocity of 275 m/s. The pressure ratio across the compressor is 12.5, the turbine inlet temperature is 1200 K, and the pressure at the exit from the nozzle is 20 kPa. Assuming an ideal jet engine and using an air standard cycle with a nominal flow rate of 1 kg/s, determine (a) the temperature and pressure at each point in the cycle, (b) the exit velocity from the nozzle, and (c) the thrust produced.

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9 Vapor Power and Refrigeration Cycles

Chapter Objectives •

To describe the operation of the steam power plant and its component parts.

•

To introduce the ideal Rankine cycle.

To show how the performance of the Rankine cycle can be enhanced through the use of superheat, reheat, and regeneration. • To examine the effects of irreversibilities. •

•

To describe the operation of the ideal vapor compression refrigeration cycle and its many component parts.

•

To indicate the operational differences between a refrigerator and a heat pump.

•

To define what is meant by a ton of refrigeration and to present the basis for refrigerant selection.

9.1

Introduction

In this chapter, we devote our attention to vapor power and refrigeration cycles. The power cycle that we consider is the Rankine or steam power plant cycle, which accounts for most of the electric power generation worldwide, far outweighing the uses of hydroelectric, nuclear, and solar power. In the Rankine cycle, the working medium is usually water, which is alternately evaporated and condensed. This is in contrast to the gas power systems discussed in Chapter 8 where the working medium is a gas that never changes phase. We also consider the vapor compression refrigeration cycle that may be found in many commercial, industrial, and household refrigeration and air-conditioning units. In the vapor compression refrigeration cycle, the working medium is the refrigerant that is alternately evaporated and condensed. We first turn to a description of the steam power plant.

9.2

The Steam Power Plant

Figure 9.1 shows the principal components of a steam power plant. The prime mover is the turbine, which takes high-pressure and high-temperature steam at the outlet of the steam generator (point 3) and discharges steam to the condenser at low pressure (point 4). The 261

262

Introduction to Thermal and Fluid Engineering Products of Combustion Steam Generator

Generator

3 Turbine Fuel Air

4 Condenser 2 Cold Water

1 Feed Water Pump

Circulating Water Pump

FIGURE 9.1 The basic components of a steam power plant.

work done by the turbine which, as we know, is a rotating device consisting of many stages, is delivered by the turbine shaft and is used to drive the generator that provides the electric power. The condenser pressure is made as low as possible because the thermal efficiency of the entire power plant is inversely proportional to the condenser pressure. The feedwater pump, operating between points 1 and 2, is used to raise the pressure of the water in the sump of the condenser to the steam generator pressure. The water leaving the pump is in the subcooled or compressed liquid region. A detail of a modern central station steam generator is displayed in Figure 9.2. In a nonnuclear power plant, the steam generator may be fired by coal (in most cases pulverized), fuel oil, or natural gas. The steam generator consists of a furnace where the steam is produced by radiation from the high-temperature products of combustion in what is frequently referred to as the waterwall. Essentially, the waterwall acts as a boiler although in some steam generators, there is a boiler bank consisting of tubes that are exposed to the products of combustion and operate in the convection mode. Often, there is an economizer consisting of a bank of finned tubes whose purpose is to extract some of the energy contained in the flue gases to heat the water entering the steam generator. The outlet of the steam generator provides steam to the turbine at point 3 in the cycle (Figure 9.1). The steam may be saturated or contain liquid water. However, because turbines using saturated or wet steam have operational problems, such as turbine blade erosion, in most cases, the steam generator contains a superheater to put the steam at point 3 in the superheated region. In many cases, there are two stages in the turbine and after the steam expands through the first stage, it is led back to the steam generator so that it can be heated once again in the reheater. The condenser receives the steam at the exhaust of the turbine at point 4 in Figure 9.1 and discharges the heat contained to the ultimate sink of circulating cooling water taken from a river or a lake. Sometimes, because of municipal restrictions, air is employed as the ultimate sink in which case the device operating between point 4 and point 1 is called a cooling tower. The pressure on the steam side of the condenser is set by the temperature of the cooling water and this temperature is consistent with the temperature difference maintained between the steam and the cooling water.

Vapor Power and Refrigeration Cycles

263

To Stack

Air In

Superheater Air Heater

Economizer

29.26 M

Air Out

8.38 M

7.42 M

Pulverizer

FIGURE 9.2 A typical central station steam generator.

9.3

The Ideal Rankine Cycle

9.3.1 The Ideal Rankine Cycle The T-s diagram for the ideal Rankine cycle is indicated in Figure 9.3 and we observe that the cycle is composed of four ideal processes: •

Process 1-2: An isentropic compression in the pump from the condenser pressure to the boiler pressure.

264

Introduction to Thermal and Fluid Engineering T

3

2 1

4

s FIGURE 9.3 The T-s diagram for the Rankine cycle. In some cases, a steam engine may take the place of the steam turbine between points 3 and 4 because of the adverse effect of moist or saturated steam on the turbine blades. •

Process 2-3: A constant pressure heat addition in the steam generator (and its associated components such as economizers and feedwater heaters) until saturated steam at the boiler pressure is obtained.

•

Process 3-4: An isentropic expansion in the turbine from the pump outlet pressure to the condenser pressure.

•

Process 4-1: A constant pressure heat rejection in the condenser.

We note in Figure 9.3 that, in the basic ideal Rankine cycle, steam leaves the steam generator and enters the turbine as saturated steam. If all work and heat interactions are considered as positive quantities, then the energy rate balance for a control volume representing just the turbine operating between state points 3 and 4 will be ˙ turb = m(h W ˙ 3 − h4)

(9.1)

where m ˙ is the mass flow rate. Equation 9.1 presumes that there is no heat transfer to the surroundings through the casing of the turbine and that there are no kinetic and potential energy changes. With the pump taken as a control volume and with operation between state points 1 and 2 in Figure 9.3, a similar analysis, again with no heat transfer and negligible kinetic and potential energy changes, will yield ˙ pump = m(h W ˙ 2 − h1)

(9.2)

The net work in the cycle is therefore ˙ cycle = W ˙ turb − W ˙ pump = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )]

(9.3)

Next, with the steam generator taken as a control volume, we note that there is no work interaction between it and its surroundings and with negligible kinetic and potential energy changes, an energy rate balance yields ˙ in = m(h Q ˙ 3 − h2)

(9.4)

The condenser has a condensing side and a cooling water side and we may apply a control volume to just the condensing side. Again, with no work interaction and with negligible

Vapor Power and Refrigeration Cycles

265

kinetic and potential energy changes, the heat rejected by the condenser will be ˙ out = m(h Q ˙ 4 − h1)

(9.5)

and because the net work of the cycle is equal to the difference between the heat added and the heat rejected ˙ cycle = Q ˙ in − Q ˙ out = m[(h W ˙ 3 − h 2 ) − (h 4 − h 1 )] or, after a slight modification ˙ cycle = m[(h W ˙ 3 − h 4 ) − (h 2 − h 1 )] which confirms Equation 9.3. The thermal efficiency of the ideal Rankine cycle is =

˙ cycle W m[(h ˙ 3 − h 4 ) − (h 2 − h 1 )] = ˙ in m(h ˙ 3 − h2) Q

or =1−

h4 − h1 h3 − h2

(9.6)

As in the case of the gas turbine, a useful parameter is the back work ratio, defined as the ratio of the power required to drive the pump to the power developed by the turbine: bwr =

˙ pump W h2 − h1 = ˙ h3 − h4 Wturb

(9.7)

We now turn to an example that is intended to illustrate the foregoing principles. The example involves saturated steam at the outlet of the steam generator and, because saturated steam at the inlet to a power plant turbine leads to operational difficulties, the example employs a reciprocating steam engine to drive the generator. The use of the steam engine does not alter, in any manner, the equations that have been developed for the steam turbine.

Example 9.1 A steam power plant operates with saturated steam at 6 MPa at the steam generator outlet and has a condenser pressure of 8 kPa. The plant employs a steam engine that can operate with saturated steam at its inlet. The mass flow rate to the plant is 25 kg/s and the cooling water for the condenser enters at 23◦ C and leaves at 31◦ C. For the ideal Rankine cycle, determine (a) the power developed by the engine, (b) the power required to drive the pump, (c) the thermal efficiency of the cycle, (d) the back work ratio, (e) the heat rejected by the condenser, and (f) the mass flow rate of the cooling water.

Solution Assumptions and Specifications (1) The water-steam mixture contained in the component parts and the interconnecting piping forms a closed system. (2) The system operates in the steady state. (3) Each component part of the system can be treated as a control volume.

266

Introduction to Thermal and Fluid Engineering T

6 MPa

3

2 1

8 kPa 4

s FIGURE 9.4 The T-s diagram for Example 9.1.

(4) This is an ideal Rankine cycle with two isentropic and two constant pressure processes. (5) Kinetic and potential energy changes are negligible. (6) The equations that involve work apply to a steam engine. (7) The condition of saturated steam at the inlet to the engine is specified. We will need to apply Equations 9.2 and 9.4 through 9.7 to satisfy the requirements of this problem. Thus, we will need specific enthalpy values at points 1 through 4 in Figure 9.4. At point 1 at P1 = 8 kPa, Table A.4 in Appendix A gives v1 = 1.008 × 10−3 m3 /kg

and

h 1 = 173.8 kJ/kg

Then, by Equation 4.10 with P2 − P1 = 6 × 106 N/m2 − 8 × 103 N/m2 = 5.992 × 106 N/m2 we find h 2 h 2 = h 1 + v1 ( P2 − P1 ) = 173.8 kJ/kg + (1.008 × 10−3 m3 /kg)(5.992 × 103 kN/m2 ) = 173.8 kJ/kg + 6.04 kJ/kg = 179.84 kJ/kg At point 3, saturated vapor at 6 MPa has specific enthalpy and entropy values that we may find in Table A.4 h 3 = 2790.5 kJ/kg

and s3 = 5.8906 kJ/kg-K

Then, at point 4, because of the isentropic expansion through the turbine, s3 = s4 = 5.8906 kJ/kg-K, Table A.4 gives at 8 kPa sf = 0.5924 kJ/kg-K

and

sfg = 7.6372 kJ/kg-K

and h f = 173.8 kJ/kg and

h fg = 2403.6 kJ/kg

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267

Then the quality at point 4, via Equation 4.8, will be x4 =

s4 − sf 5.8906 kJ/kg-K − 0.5924 kJ/kg-K = = 0.6937 sfg 7.6372 kJ/kg-K

Now we can find h 4 using Equation 4.6b h 4 = h f + xh fg = 173.8 kJ/kg + (0.6937)(2403.6 kJ/kg) = 173.8 kJ/kg + 1667.4 kJ/kg = 1841.2 kJ/kg We now proceed to the determination of the required quantities. (a) The power developed by the engine is given by Equation 9.1. With m ˙ = 25 kg/s ˙ engr = m(h W ˙ 3 − h4) = (25 kg/s)(2790.5 kJ/kg − 1841.2 kJ/kg) = (25 kg/s)(949.3 kJ/kg) = 23,732 kW ⇐ (b) The power required to drive the pump is given by Equation 9.2. ˙ pump = m(h W ˙ 2 − h1) = (25 kg/s)(179.84 kJ/kg − 173.8 kJ/kg) = (25 kg/s)(6.04 kJ/kg) = 151 kW ⇐ (c) The thermal efficiency of the cycle is given by Equation 9.6.  = 1− = 1−

h4 − h1 1841.2 kJ/kg − 173.8 kJ/kg =1− h3 − h2 2790.5 kJ/kg − 179.84 kJ/kg 1667.4 kJ/kg = 1 − 0.639 = 0.361 ⇐ 2610.7 kJ/kg

˙ eng substituted for W ˙ turb . (d) The back work ratio is given by Equation 9.7 with W bwr =

˙ pump W 151 kW = = 0.0064 ˙ 23,732 kW Weng

or

0.64% ⇐

(e) We find the heat rejected in the condenser from Equation 9.5. ˙ out = m(h Q ˙ 4 − h1) = (25 kg/s)(1841.2 kJ/kg − 173.8 kJ/kg) = (25 kg/s)(1667.4 kJ/kg) = 41, 685 kW ⇐ (f) If the water side of the condenser is considered as a control volume, we see that, in the absence of any work interactions with the surroundings and negligible kinetic and potential energy changes, an energy rate balance gives ˙ out = m Q ˙ w (h 2w − h 1w ) where the subscript, w, refers to the cooling water. Table A.3 may be used to find, with interpolation, at T1 = 23◦ C and T2 = 31◦ C, respectively h 1w = 96.44 kJ/kg

and

h 2w = 129.78 kJ/kg-K

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Introduction to Thermal and Fluid Engineering

With the heat transferred to the cooling water from part (e) m ˙w = =

˙ out Q 41,685 kW = h 2w − h 1w 129.78 kJ/kg-K − 96.44 kJ/kg-K 41,685 kW = 1250.3 kg/s ⇐ 33.34 kJ/kg-K

9.3.2 Increasing the Efficiency of the Ideal Rankine Cycle In Example 9.1, we observed that the efficiency of the ideal Rankine cycle was a rather dismal 0.361. This leads us to the conclusion that the increase in the efficiency of the cycle is a subject of more than academic interest. We will find it extremely helpful if we can develop an expression that is similar in form to the equation for the thermal efficiency of the Carnot cycle given in Chapter 6 by =1−

TL TH

In Figure 9.5, we know that heat is added between points 2 and 3 and is represented, except for the nonisothermal heat supply in the compressed liquid region, by the integral  q in =

s3

Tds s2

Because T is a function of s, we can obtain its average value, T¯ H , from T¯ H =

1 s3 − s2



s3

Tds s2

and thus, q in = T¯ H (s3 − s2 ) T

TH

2´

3

2 1

1´

4 TL

s FIGURE 9.5 A T-s diagram showing the comparison between the ideal Rankine cycle and the Carnot cycle.

Vapor Power and Refrigeration Cycles

269

When we proceed in identical fashion, we can show that  s4 q out = Tds s1

with an average value such that q out = T¯ L (s4 − s1 ) Hence, with s1 = s2 and s3 = s4 , the work of the cycle wcycle = q in − q out can be written as wcycle = T¯ H (s3 − s2 ) − T¯ L (s3 − s2 ) so that the efficiency will be =1−

T¯ L T¯ H

(9.8)

Equation 9.8 suggests that the efficiency of the ideal Rankine cycle is improved as T¯ L is minimized or as T¯ H is maximized—or both. Except that average rather than fixed hot and cold temperatures are employed, the form of Equation 9.8 is similar to the form of the equation for the Carnot cycle efficiency. It is a fact, which we can verify by an inspection of the steam tables, that as the saturation pressure is increased, the saturation temperature is increased. Thus, we find that higher ideal Rankine cycle efficiencies occur at higher steam generator pressures. By similar observations, we know that higher ideal Rankine cycle efficiencies occur at lower condenser pressures. There are, of course, limitations to these conclusions. The convecting surfaces in the steam generator and condenser are heat transfer devices that require a temperature difference to yield a flow of heat. The size of the steam generator is inversely proportional to the temperature difference between the flue gases and the steam. Moreover, as we proceed through the steam generator, we find that the maximum temperature of the steam at a point in the cycle cannot exceed the lowest flue gas temperature at that point. At the condenser end, the temperature is not only limited by the temperature of the cooling medium at the exit but also by the desire to operate over a temperature difference that will allow for a condenser of reasonable size. This entire subject is treated in Chapter 25.

9.4

The Ideal Rankine Cycle with Superheat

A superheater is essentially a heat exchanger that can be placed in the furnace of the steam generator so that it can absorb radiant heat from the flue gases. Alternatively, the superheater may consist of a bank of tubes placed in such a manner that the flue gases pass across it and transfer heat to it by convection. In either application, the purpose of the superheater is to raise the temperature of the steam leaving the steam generator and entering the turbine. A T-s diagram for the ideal Rankine cycle with superheat is shown in Figure 9.6. Because the temperature at point 3 in the cycle of Figure 9.6 is elevated above the saturation temperature, the average temperature, TH , has increased. In accordance with

270

Introduction to Thermal and Fluid Engineering T

3

2 1

4 s

FIGURE 9.6 The T-s diagram for the ideal Rankine cycle with superheat.

Equation 9.8, we should see an increase in efficiency. Moreover, we can see that the use of the superheater begins to alleviate the problem of low-quality steam at the turbine exhaust. Here too, state points 1 through 4 are used for the ideal Rankine cycle with superheat. Thus, Equation 9.1 through 9.7 may be used when the superheater is employed. However, we will find in Example 9.2, which now follows, that the values of specific enthalpy and entropy will change. We will also see in Example 9.2 how the performance changes when the superheater is in the cycle.

Example 9.2 A steam power plant operates with superheated steam at 6 MPa and a temperature of 600◦ C at the steam generator outlet and has a condenser pressure of 8 kPa (Figure 9.7). The mass flow rate in the plant is 25 kg/s and the cooling water for the condenser enters at 23◦ C and leaves at 31◦ C. For the ideal Rankine cycle, determine (a) the power developed by the turbine, (b) the power required to drive the pump, (c) the thermal efficiency of the cycle, (d) the back work ratio, (e) the heat rejected by the condenser, and (f) the mass flow rate of the cooling water.

T

4 600°C 6 MPa

2 1

8 kPa 4

s FIGURE 9.7 The T-s diagram for Example 9.2.

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Solution We make the same assumptions as in Example 9.1 except that a turbine is now involved with its inlet taking superheated steam and conditions at state points 1 and 2 remain the same. Conditions at state points 1 and 2 are the same as those in Example 9.1. h 1 = 173.8 kJ/kg

and

h 2 = 179.84 kJ/kg

At point 3, superheated vapor at 6 MPa and 600◦ C has specific enthalpy and entropy values that we may find in Table A.5 h 3 = 3654.2 kJ/kg

and s3 = 7.1627 kJ/kg-K

Then, at point 4, because of the isentropic expansion through the turbine, s3 = s4 = 7.1627 kJ/kg-K, Table A.4 gives at 8 kPa s f = 0.5924 kJ/kg-K

and sfg = 7.6372 kJ/kg-K

and h f = 173.8 kJ/kg

and

h fg = 2403.6 kJ/kg

We find the quality at point 4 via Equation 4.8: x4 = =

s4 − s f 7.1627 kJ/kg-K − 0.5924 kJ/kg-K = sfg 7.6372 kJ/kg-K 6.5703 kJ/kg-K = 0.8603 7.6372 kJ/kg-K

and then by Equation 4.6b h 4 = h f + xh fg = 173.8 kJ/kg + (0.8603)(2403.6 kJ/kg) = 173.8 kJ/kg + 2067.8 kJ/kg = 2241.6 kJ/kg We now proceed to the determination of the required quantities. (a) The power developed by the turbine is given by Equation 9.1. With m ˙ = 25 kg/s ˙ turb = m(h W ˙ 3 − h4) = (25 kg/s)(3654.2 kJ/kg − 2241.6 kJ/kg) = (25 kg/s)(1412.6 kJ/kg) = 35,315 kW ⇐ (b) The power required to drive the pump is given by Equation 9.2 and has the same value as in Example 9.1 151 kW ⇐ (c) The thermal efficiency of the cycle is given by Equation 9.6  = 1− = 1−

h4 − h1 2241.6 kJ/kg − 173.8 kJ/kg =1− h3 − h2 3654.2 kJ/kg − 179.84 kJ/kg 2067.8 kJ/kg = 1 − 0.595 = 0.405 ⇐ 3474.4 kJ/kg

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(d) The back work ratio is given by Equation 9.7 bwr =

˙ pump W 151 kW = = 0.0043 ˙ 35,315 kW Wturb

or

0.43% ⇐

(e) We find the heat rejected in the condenser from Equation 9.5 ˙ out = m(h Q ˙ 4 − h1) = (25 kg/s)(2241.6 kJ/kg − 173.8 kJ/kg) = (25 kg/s)(2067.8 kJ/kg) = 51, 695 kW ⇐ (f) The enthalpy values for the cooling water remain the same as in Example 9.1. We find that at T1 = 23◦ C and T2 = 31◦ C, respectively h 1w = 96.44 kJ/kg

and

h 2w = 129.78 kJ/kg-K

With the heat transferred to the cooling water from part (e) ˙ out = m Q ˙ w (h 2w − h 2w ) and m ˙w = =

˙ out Q 51, 695 kW = h 2w − h 1w 129.78 kJ/kg-K − 96.44 kJ/kg-K 51,695 kW = 1550.5 kg/s ⇐ 16.66 kJ/kg-K

This example shows a modest increase in efficiency over the value obtained in Example 9.1. The primary reason for superheat is to increase the quality of the steam-water mixture at the low-pressure end of the turbine.

9.5

The Effect of Irreversibilities

Several irreversibilities conspire to reduce the thermal efficiency of the Rankine cycle. The most significant of these is the departure from the isentropic compression in the pump and the isentropic expansion in the turbine. As shown in Figure 9.8, these compressions and expansions are not isentropic and the specific enthalpies at the pump and turbine outlets are at higher values than in the ideal cycle. We have shown in Chapter 7 that irreversibilities in the pump can be handled by a parameter known as the isentropic pump efficiency s =

h 2s − h 1 h2 − h1

(7.33)

where h 2s is the specific enthalpy of the water at the pump outlet when the compression in the pump is isentropic and h 1 is the specific enthalpy of the water at the pump inlet. If the isentropic pump efficiency and specific enthalpy values (including h 2s ) for the ideal

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T 3

2 2s 4 1

4s

s FIGURE 9.8 The T-s diagram for the ideal Rankine cycle modified to show the effect of irreversibilities in the pump and turbine.

Rankine cycle are known or obtained from the steam tables, the specific enthalpy, h 2 may be computed from h2 = h1 +

h 2s − h 1 s

(9.9)

An identical procedure may be followed for the turbine. In Chapter 7, we showed that irreversibilities in the turbine are accounted for by the isentropic turbine efficiency.1 s =

h3 − h4 h 3 − h 4s

(7.32)

where h 4s is the specific enthalpy of the steam (or steam-water mixture) at the turbine outlet when the turbine irreversibility is excluded and h 3 represents the specific enthalpy of the steam at the turbine inlet. If the isentropic turbine efficiency and the specific enthalpy values (including h 4s ) for the ideal Rankine cycle are known or obtained from the steam tables, the specific enthalpy, h 4 may be calculated from h 4 = h 3 − s (h 3 − h 4s )

(9.10)

Other irreversibilities manifest themselves in pressure losses in the tubing in the steam generator and condenser and the connecting pipes between the power plant components. There is also an unwanted heat transfer through the casing of the various power plant components to the environment. However, these additional irreversibilities will not be considered here. We now reconsider Example 9.2 and show how the inclusion of effects of pump and turbine irreversibilities influence thermal efficiency.

Example 9.3 Consider the Rankine cycle of Example 9.2 and suppose that both the pump and the turbine have isentropic efficiencies of 86% (0.860). All other conditions are the same as in Example 9.2. Determine the thermal efficiency of the cycle that includes both the pump and turbine irreversibilities (Figure 9.9). 1 The

subscripts have been adjusted in keeping with Figure 9.7.

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Introduction to Thermal and Fluid Engineering T

3 6 MPa

600°C

2 2s 8 kPa 1

4s

4

s FIGURE 9.9 The T-s diagram for Example 9.3.

Solution We make the same assumptions as in Example 9.2 except that the pump and turbine do not include isentropic processes. Specific enthalpy data at each state point in the ideal Rankine cycle may be taken from Example 9.2 but we must designate the specific enthalpies at points 2 and 4 as h 2s and h 4s , which are the constant entropy values h 1 = 173.8 kJ/kg h 2s = 179.84 kJ/kg

h 3 = 3654.2 kJ/kg h 4s = 2241.6 kJ/kg

Now we use Equations 9.9 and 9.10 to determine the actual values of h 2 and h 4 . For h 2 , Equation 9.9 gives h2 = h1 +

h 2s − h 1 179.84 kJ/kg − 173.8 kJ/kg = 173.8 kJ/kg + pump 0.860

= 173.8 kJ/kg +

6.04 kJ/kg 0.860

= 173.8 kJ/kg + 7.02 kJ/kg = 180.82 kJ/kg and for h 4 , Equation 9.10 provides h 4 = h 3 − turb (h 3 − h 4s ) = 3654.2 kJ/kg − (0.860)(3654.2 kJ/kg − 2241.6 kJ/kg) = 3654.2 kJ/kg − (0.860)(1412.6 kJ/kg) = 3654.2 kJ/kg − 1214.84 kJ/kg = 2439.4 kJ/kg We now have a different set of specific enthalpies some of which account for the irreversibilities h 1 = 173.8 kJ/kg

h 3 = 3654.2 kJ/kg

h 2 = 180.82 kJ/kg

h 4 = 2439.4 kJ/kg

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The thermal efficency of the cycle is now  = 1− = 1−

h4 − h1 2439.4 kJ/kg) − 173.8 kJ/kg =1− h3 − h2 3654.2 kJ/kg) − 180.82 kJ/kg 2265.6 kJ/kg = 1 − 0.652 = 0.348 ⇐ 3473.4 kJ/kg

It is apparent that a lack of attention to the irreversibilities in the pump and turbine (and particularly the turbine) can lead to an enormous disappointment.

9.6

The Rankine Cycle with Superheat and Reheat

In the ideal Rankine cycle with both reheat and superheat (Figure 9.10a), the turbine has two stages. After an expansion in stage 1, the high-pressure turbine, the steam is rerouted to the steam generator where it is preheated at constant pressure in a reheater. The steam Products of Combustion Turbine Stage 1

Turbine Stage 2 Generator 6

3 Reheater Fuel Air

5 2

Pump

4

Hot Water

Condenser

Cold Water

1 Pump (a)

T 3 6

5

2 1 4 (b)

s

FIGURE 9.10 (a) A steam power plant with both superheat and reheat. (b) The T-s diagram for the ideal Rankine cycle.

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is then led to stage 2, called low-pressure turbine where it is expanded to the condenser pressure. In Figure 9.10b, the T-s diagram for the ideal Rankine cycle with both superheat and reheat, we see that there are six state points. Points 1 through 4 are in their customary location. However, point 5 is the point where the steam is taken from the high-pressure turbine and led back to the reheater inside of the steam generator. Point 6 is located at the end of the constant pressure reheat process and is at the inlet to the low-pressure turbine. At point 4, where the steam is exhausted from the low-pressure turbine, we see that the steam may still be superheated or possess a high quality. This is a significant advantage of the reheat cycle. Whether the overall thermal efficiency of the power plant operating with reheat is improved depends on the selection of the reheat pressure and the temperature at the end of the reheat process. State 5 is usually permitted to come close to the saturated vapor line and the temperature at the end of the reheat process, T6 , is equal to or somewhat less than the total steam temperature, T3 . Use of the reheat cycle with 0.200
t2 > t1

Steady State

t3

Stationary Surface FIGURE 11.3 Velocity profiles between parallel plates as a function of time between initial motion and steady state.

If this is divided by the molecular weight of the gas (or gas mixture) in question, the equation takes the form Pv =

¯ R T = RT M

(11.2)

The form of this equation of state that we will find most useful is P = RT

(11.3)

where  is the mass density of the gas in kg/m3 . Fluid properties of air and water are provided in Tables A.13 and A.14 in Appendix A. The concept of flow properties is not as clear as that of fluid properties. The general idea is that certain fluid behavior is a characteristic of the flow condition rather than being specific for a given fluid. An example of a flow property is velocity. Some others will be considered shortly. However, certain fundamental concepts must be discussed before these additional ideas can be introduced.

11.3

The Variation of Properties in a Fluid

If we consider the property, pressure, which we have designated as P, and its variation with position in a two-dimensional framework, we may write P = P(x, y)

(11.4)

A familiar example of pressure variation is a weather map, which is commonly found in your daily newspaper. The lines on such a map (Figure 11.4) are lines of constant pressure called isobars. The prefix iso- means constant and examples of its use are in isothermal for constant temperature, isometric for constant measure, and isotropic for constant properties in all directions. In Figure 11.5, it is of interest to evaluate the change in pressure as we move from one position at P1 (x1 , y1 ) to another at P2 (x2 , y2 ). The change in P can be determined by expressing its differential in the form dP =

∂P ∂x

dx +

∂P ∂y

dy

(11.5)

Integration of this expression between the two points of interest will allow the pressure difference, P2 − P1 , to be evaluated. To do this, we must know the partial derivatives ∂ P/∂x and ∂ P/∂ y.

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1026

1008 6 1010 1012 8

1024

6 1022

10

6

1020 1014 8

10

8

10

1016

H 1022 8

40 38

6

1014

1010 1012

36 34

1020

32 30 28

8

1018

1012

26

8

24

1016

22

10

20 6

18

1014

16 8

14 8

1016

L 1010

10 12 14 12 10

?? 1012

8

Observed Sea Level Pressure (mb) 030722/2000 UTC Wind Speed (kt) 030722/2000 UTC FIGURE 11.4 A weather map showing isobars. (From Ohio State University Weather Service, http:/aspl.ohio-state.edu).

It will also be of interest to evaluate the rate of pressure change along a given path. Denoting a general path or direction as s, we express the directional derivative as dP ∂ P dx ∂ P dy = + ds ∂ x ds ∂ y ds

(11.6)

There are, of course, an infinite number of paths that might be taken in moving away from a particular point. Two such paths are of particular interest. One is the path along which dP/ds is a minimum. Obviously this is dP/ds = 0 leading to P = C (a constant) and is the path along an isobar. The other is the path along which dP/ds is a maximum.

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333

y

P2

ds

dy θ dx

P1

x FIGURE 11.5 Isobars in the x-y plane.

The directional derivative along the path of maximum rate of change is termed the gradient designated by grad or ∇. The gradient of the pressure in a two-dimensional frame is written as ∇P =

∂P ∂x

∂P

x+

∂y

(11.7)

y

where x and y are unit vectors in the x- and y-directions, respectively. The definition of the gradient can be extended to three dimensions. In rectangular coordinates with P = P(x, y, z), we have ∇P =

∂P ∂x

x+

∂P ∂y

y+

∂P ∂z

z

(11.8)

The expressions for ∇ P in cylindrical and spherical coordinates are given in Appendix B. The reader will observe that, while these expressions are different, the geometric meaning of the gradient, that is, the vector having direction and magnitude of maximum rate of change of the dependent variable, remains the same.

Example 11.1 For a scalar  = xy3 z2 , determine grad  and evaluate it at the point x = 1, y = 1/2, z = 1/4.

Solution

Here with  = xy3 z2 ∂ ∂x ∂ ∂y ∂ ∂z

= y3 z2 = 3xy2 z2 = 2xy3 z

Thus, ∇ = y3 z2 x + 3xy2 z2 y + 2xy3 zz ⇐

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and at the point (x = 1, y = 1/2, z = 1/4)  3  2  2  2  3   1 1 1 1 1 1 ∇ = x + 3(1) y + 2(1) z 2 4 2 4 2 4       1 3 1 = x+ y+ z 128 64 16 =

 1  x + 6y + 8z ⇐ 128

The path of maximum variation in pressure is perpendicular to the path of minimum or zero variation. We thus deduce that, to move in a direction of maximum rate of pressure change, we should proceed along a path that is at right angles to an isobar. This relationship between isolines and directions of maximum rate of change is general.

11.4

The Continuum Concept

In our discussion thus far, it has been presumed that the pressure varies continuously in any direction. Another way of stating this idea is that ∂ P/∂x and ∂ P/∂ y have values at the point (x, y). Such variation requires the existence of a continuum. To develop this concept, we will consider the density, . We already know that density is the mass per unit volume. Beyond that, of course, we must consider just what we mean when we refer to the density at a point. If we consider the mass of air in, say, a cubic meter of space, we may evaluate the density according to =

m V

(11.9)

where V is the volume (1 m3 in this case) and m is the mass contained in V. We next consider a volume V that is considerably smaller than 1 m3 , say a cube measuring 1 cm on a side. In this case, the density expression would be the same as Equation 11.9 and we would expect the same result for . We now continue the procedure, evaluating m/V for ever smaller values of V. A plot of this process is shown in Figure 11.6. Eventually, the volume becomes sufficiently Molecular Regime Δm ΔV

Continuum Regime

δV FIGURE 11.6 The density at a point.

ΔV

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small so that it will contain different numbers of molecules at different times of observation. From this point, as V becomes smaller, the variation in m/V oscillates between ever larger extremes. Clearly, the extreme values on this ratio become very large in the limit as V −→ 0. For V greater than a limiting value, V,  is constant and we call this the continuum regime. When V is less than V, the density, , will vary and this is the molecular regime. When we consider the density at a point to behave continuously, that is, just as though continuum conditions prevail, we are not being precise. However, such operations are still physically realistic because V, for all practical purposes, is extremely small. Operationally, properties will be assumed to be continuous quantities. The density at a point will actually be considered to hold for a small region, V, near a point such that =

lim

V−→V

m V

(11.10)

The concept of a continuum will be considered to hold throughout this book. Point-to-point continuous behavior will thus be presumed for all fluid and flow properties.

11.5

Laminar and Turbulent Flow

The terms laminar and turbulent flow are familiar to most of us and, intuitively, we have an impression as to the meaning of these concepts. A familiar situation is shown in Figure 11.7 where the characteristics of laminar and turbulent flows can be clearly seen. In Figure 11.7, smoke rises from a source of combustion in an apparently smooth, regular fashion and then suddenly changes to a much less regular and chaotic flow. Smooth, orderly flow is called laminar. The term literally means layer-like and it implies that fluid flows in distinct layers, or “lamina,” which retain their identity. In such a condition, fluid particles in a given layer remain in that layer as long as laminar flow continues. The only migration of fluid between adjacent layers is at the molecular level. Turbulent flow

Laminar flow

FIGURE 11.7 A common example of the transition from laminar to turbulent flow.

336

Introduction to Thermal and Fluid Engineering Turbulent Eddies

y Average Velocity Profile A

A x V

FIGURE 11.8 The structure of turbulent flow.

In contrast to the regularity of laminar flow, fluids in turbulent flow are characterized by apparently random motions of large numbers of fluid particles in directions other than that of the mean flow. Fluid particles that start at one position in a turbulently flowing stream may be carried, along with many neighbors, into another location by a small whirlpool-like eddy. The presence of eddies is a characteristic of turbulent flow and, even though the main flow is in the x-direction in Figure 11.8, turbulent eddies will propel fluid particles randomly between different y locations. As we might expect, laminar flow is much easier to describe quantitively than is turbulent flow. We will consider both types of flow in subsequent chapters. The region where the flow changes from laminar to turbulent, as seen in Figure 11.7, is designated as the transition regime.

11.6

Fluid Stress Conventions and Concepts

We defined a fluid in Section 11.1 in terms of the response to an imposed shearing stress. In a large quantity of fluid, an individual element may have imposed on it a number of influences, in addition to the shear stress. Forces exerted on a medium are of two types. Body forces are those developed without physical contact and they arise from the presence of the fluid in a force field; examples would be gravitational and electromagnetic fields. The gravitational force is the most common type of body force and will be the only such force considered in this book. Quantitatively, Newton’s second law of motion describes the gravitational force as F = W = Vg

(11.11)

where  is the mass density of the fluid, V is the volume under consideration and g is the acceleration of gravity. We commonly call such a force the weight, W, of the fluid volume. Various forms of Equation 11.11 give rise to other terms; W/V, the weight per unit volume is the specific weight, , and g may be thought of as the gravitational body force per unit mass. It should be noted that Equation 11.11 is a vector expression and that the direction of the body force, F, must be the same as that of, g, the gravitational acceleration. If gravity is

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oriented downward (in the minus y-direction), we may write Fx x + F y y + Fz z = V(−gy) giving F x = Fz = 0

(11.12a)

F y = W = −Vg

(11.12b)

and

Expressions for the weight will generally have obvious directional characteristics so we will not, henceforth, evaluate W with this much formality. Surface forces are those exerted by direct contact between a fluid medium and a boundary and the result of forces acting on a fluid medium is to put a general element of the fluid into a state of stress. The variation of the stress throughout the fluid is termed the stress field. A piece of the boundary of a general fluid element is shown in Figure 11.9. We note that a small portion of the surface, S, has a surface force, F acting on it. The vector, n, is the outwardly directed unit vector normal to S. The force, F, can thus be resolved into its normal and tangential components, Fn and Ft , as indicated. We then define the normal and shear stresses at S, consistent with continuum behavior as  = lim

Fn S

(11.13)

 = lim

Ft S

(11.14)

S−→0

and S−→0

respectively. These stress components,  and , will have varying character depending on the orientation of the surface, S, used in their definitions. Stress is, in general, a tensor quantity of second order: this means that its complete specification involves nine components. This is in contrast to a vector quantity (a firstorder tensor) having three components and a scalar (a zeroth-order tensor), which has only one. In Figure 11.10, we show two elements in a fluid continuum. The configuration of the element in Figure 11.10a is consistent with Cartesian or rectangular coordinates; its dimensions are x, y, and z in the directions corresponding to the coordinate directions.

ΔFn ΔF n ΔS ΔFt

FIGURE 11.9 A surface force acting on an element of surface S.

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Introduction to Thermal and Fluid Engineering σyy

τyx τyz τxy τzy

σxx

Δy

τzx

τxz

σzz Δz y

Δx x

z (a) σzz τzr τzθ Δz τrθ z θ

σrr r

τrz rΔθ

τθr τθz Δr

σθθ

(b) FIGURE 11.10 Stress in fluid elements: (a) rectangular coordinates and (b) cylindrical coordinates.

In Figure 11.10b, we show an element consistent with a cylindrical coordinate system where we note that its sides have lengths r, r , and z. On each face, the stress can be resolved into three components as shown. By conventions, shear stresses, parallel to the face, are labeled  and normal stresses, perpendicular to the face, are labeled . Because each such element has six faces, there are six sets of stress components and the three sets on the visible faces are shown in the figure. As mentioned earlier in this section, these nine components are conveniently expressed as a matrix ⎡

⎤ xx xy xz ⎢ ⎥ ⎣  yx  yy  yz ⎦ zx zy zz

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339 σyy τyx τxy

σxx

σxx

Δy

y Δx τxy τyx

x

σyy (a) σzz τzr

τrz σrr

σrr

Δz

z

Δr τrz τzr z σzz (b)

FIGURE 11.11 Two-dimensional views showing stress components: (a) the x-y plane and (b) the r -z plane.

where the diagonal elements, ii , are the normal stress components having both subscripts the same. The off-diagonal elements, ij , are the shear stress components and we note that the subscripts on  are mixed, that is, they are not the same. The convention for the labeling of ii and ij merits our closest attention. The first subscript identifies the plane on which the stress component acts and the second identifies the direction of action relative to the coordinate system. The plane, or face, of the element is specified by the direction that the outwardly directed normal vector to that face will point. Thus, the three faces shown in Figure 11.10a are the positive x, y, and z faces. In Figure 11.10b, these are the , r , and z faces. All components are conventionally drawn and considered in their positive sense and positive components are those having both subscripts with the same sign, either positive or negative. Two-dimensional representations of these elements are shown in Figure 11.11. The x-y plane is shown in Figure 11.11a and the r -z plane in Figure 11.11b. We will use such elements frequently in the chapters to follow.

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Introduction to Thermal and Fluid Engineering Jyx Δx

Δx Δy

Δy

θ

θ (a)

(b)

FIGURE 11.12 The deformation of a fluid element (a) at time t and (b) at time t + t.

11.7

Viscosity, a Fluid Property

We are now able to give meaning to a very important fluid property, the viscosity, symbolized by (the Greek lower case “mu”), which is defined as viscosity =

shear stress rate of shear strain

(11.15)

Figure 11.12 shows a two-dimensional view of a fluid element considered at two instants in time and shows the change of the configuration, that is, the deformation, resulting from the presence of a shearing stress. The “word equation” of Equation 11.15 may be written in terms of the quantities shown in Figure 11.12  yx = (11.16) −d/dt We recall that shear strain is related to the change in the angle . The rate of shear strain would thus be the time derivative of . Upon examining the relationship between cause (the shear stress) and effect (the angular deformation), we notice that a positive shear will cause a decrease in  and a negative shear will cause  to increase. This is the reason for the negative sign in Equation 11.16. We may evaluate d/dt in terms of the flow condition causing the deformation shown. Clearly, for  to increase in the time period, t, the velocity, Vˆ x , must be greater at y + y than at y. The rate of shearing stress is −d/dt and we proceed to express it as −

d |t+t − |t = lim x, y, t−→0 dt t

Here t =

2

and |t+t =

( Vˆ x | y+y − Vˆ x | y )t

− arctan 2 y

so that ( Vˆ x | y+y − Vˆ x | y )t

− arctan − d 2 y 2 − = lim t−→0 dt t

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Observe that the two /2 terms cancel, which allows the t’s to be cancelled as well. We then have ( Vˆ x | y+y − Vˆ x | y ) d d Vˆ x − = lim = (11.17) y−→0 dt y dy The minus sign is the result of a positive shear stress, so a consistent representation for the viscosity will be  yx = (11.18) ˆ d V x /dy and the common form of Equation 11.18 is written as  yx =

d Vˆ x dy

(11.19)

which is referred to as Newton’s viscosity relationship. Many authors refer to Equation 11.19 as Newton’s “law” of viscosity. Clearly, it is not a law in a fundamental sense so we will avoid such a designation. A fluid, having a viscosity that does not vary as a function of strain rate, is termed Newtonian. Water, all gases, and a number of fluids are Newtonian. Some other fluids, such as starch suspensions and paints, display varying values of depending on the rate of deformation. These fluids are designated non-Newtonian. Equations 11.18 and 11.19 represent viscosity in the case of one-dimensional flow. A more general designation for velocity variation in all three coordinate directions would yield the expressions

ˆy ˆx ∂V ∂V  yx = xy = − (11.20a) ∂y ∂x

ˆy ˆz ∂V ∂V zy =  yz = − (11.20b) ∂z ∂y and

xz = zx =

ˆz ∂V ∂x

−

ˆx ∂V ∂z

(11.20c)

in rectangular coordinate form. Viscosity is generally a function of temperature. In gases, the viscosity increases with temperature as a result of increased molecular activity. In contrast, the much more closely packed molecules in liquids experience decreased intermolecular forces as the temperature rises. Hence, the viscosities of liquids decrease with temperature. Viscosity is not affected by pressure except for extremes at both the high and low ends. Thus, we generally consider viscosity to be independent of pressure. Viscosity is tabulated, as a function of temperature, for a number of important fluids, along with other properties in Table A.15 for gases and A.16 for liquids. Another property, tabulated along with , is the kinematic viscosity, designated by  (the Greek lower-case “nu”), which is defined as ≡ (11.21)  Because density varies with both temperature and pressure, the kinematic viscosity will also vary with both properties.

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The units of viscosity can be determined from Equation 11.18. We have =

shear stress N/m2 N-s = = 2 = Pa-s velocity/length m/s/m m

(11.22)

or, representing F as ma , an alternate form is N-s kg-m s kg = = 2 2 2 m s m m-s

=

(11.23)

Both of these representations are correct, the form of Equation 11.23 is the one most often seen. Another term seen occasionally is the poise defined as 1 g/cm-s. Units for the kinematic viscosity are seen to be m2 /s and an additional unit is the stoke defined as 1 cm2 /s. The use of Equation 11.18 is demonstrated in Examples 11.2 and 11.3.

Example 11.2 The hydraulic ram or hoist in an automobile lift consists of a circular shaft with an outer diameter of 37.88 cm, which moves inside a stationary cylindrical housing having an inside diameter of 37.91 cm. The space between the stationary and moving surfaces is filled with hydraulic fluid having a viscosity of 0.304 kg/m-s. The desired rate of travel of the hoist is 0.18 m/s. (a) Determine the frictional force (resistance) for a 2.5-m long hoist. (b) Assume that a length of 2.5 m of the ram (the moving cylinder) is engaged and determine the rate of sink of a ram loaded with 700 kg when the downward gravitational force is balanced by the frictional force due to the viscosity of the hydraulic fluid (Figure 11.13).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluid is incompressible. (a) An expanded view of the space between the two surfaces is shown in Figure 11.13b. Because the gap thickness,  = 0.015 cm, is so small relative to the ram diameter,  0.015 cm = = 3.960 × 10−4 D 37.88 cm we will neglect any influence of the surface curvature.

h V Δ D (a)

(b)

FIGURE 11.13 Hydraulic ram for Example 11.2: (a) the configuration and (b) a detail of the space between the shaft and the housing.

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The resisting force is evaluated as   F =  d S =  D dy =  Dh S

and with the curvature neglected =

d Vˆ Vˆ = dx 

Therefore, Vˆ

Dh  (0.304 kg/m-s)(0.18 m/s)( )(0.3788 m)(2.5 m) = 1.5 × 10−4 m = 1085 N ⇐

F =

(b) With the frictional force set equal to the load F =

Vˆ

Dh = mg 

we have mg Vˆ = Dh =

(700 kg)(9.81 m/s2 )(1.5 × 10−4 m) (0.304 kg/m-s)( )(0.3788 m)(2.5 m)

= 1.14 m/s ⇐

Example 11.3 A circular disk has its plane surface parallel to a stationary plate

(Figure 11.14). The space between the two surfaces, a distance  apart, contains a fluid with a viscosity, . Develop a relationship for the torque, T, required to rotate the disk at a uniform angular velocity, .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluid in the space between the surfaces is incompressible. (3) There are no edge effects.

ω Δ D

ω D

FIGURE 11.14 Rotating disk for Example 11.3.

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ω

S r dr ds = 2πrdr FIGURE 11.15 Detail of rotating disk.

This is another application of Newton’s viscosity relationship, but in this case, because the moving surface is rotating, the velocity is a variable and is a function of the distance from the axis of rotation. For this case, the viscous force, dF, associated with the area d S, as shown in Figure 11.15, will be dF = d S = (2 r dr ) The area, d S, is chosen in this manner because it lies at a distance, r , from the axis of rotation. The velocity gradient is therefore, Vˆ r r =   where  is the separation between the disk and the plate. The problem is now solved in a straightforward manner 



T=

r dF = 

D/2

=

r

0

=

2 

r d S

r (2 r dr ) 



D/2

r 3 dr

0

or  2 r 4  D/2 D4 T = = ⇐  4 0 32 A device that is used to measure the viscosity of a gas or a liquid is called a viscometer. Such a device may take on a variety of forms and Design Example 4, which now follows, illustrates the use of a rotating cylinder as a viscometer.

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δ

ω

FIGURE 11.16 Rotating drum viscometer.

11.8

Design Example 4

One type of viscometer is constructed of two concentric cylinders separated by a very narrow gap. Such a device is shown in Figure 11.16 where the diameter of the inner (rotating) cylinder is di = 16 cm, the diameter of the outer (stationary) cylinder is do = 16.30 cm and the length of both cylinders is L = 50 cm. The gap is filled with an oil of unknown viscosity and the inner cylinder is rotated at 300 rpm. If the measured torque is 1.84 N-m, determine the viscosity of the oil.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is no flow of fluid. (3) The fluid is incompressible. (4) There are no end effects. (5) The shear stress in the gap is directly proportional to the velocity gradient. The torque necessary to produce a steady-state rotation of the inner cylinder is equal to the product of the viscous force imparted to the inner surface by the oil and the distance, ri , from the axis of rotation. The viscous shear stress is given by  ˆ  dV =  dr  r =ri

The resulting viscous force is  F = A = (2 ri L) =

 ˆ  dV  dr 

r =ri

 (2 ri L)

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and the required torque is

 T = F ri =

 ˆ  dV  dr 





2 ri2 L



r =ri

When the size of the gap, , between the rotating and stationary cylinders is relatively small (/ri  1), we may approximate the velocity gradient by the linear relationship  ˆ  ˆ ˆi − 0 dV V V ri = =  dr  r   r =ri

Our torque expression is now T =

 ri  2 ri2 L 

and the viscosity can be evaluated as =

T 2 ri3 L

and with T = 1.84 N-m

(specified)

do − di 16.30 cm − 16.00 cm = = 0.15 cm 2 2 16.00 cm ri = = 8.00 cm 2 L = 50 cm =

and =

300 rev/min (2 rad/rev) = 10 rad/s 60 s/min

we have =

11.9

(1.84 N-m)(0.0015 m) = 0.0546 kg/m-s ⇐ 2 (0.080 m) 3 (0.50 m)(10 rad/s)

Other Fluid Properties

11.9.1 Specific Gravity and Specific Weight Density, , specific volume, v, specific gravity, SG, and specific weight, , were considered in detail in Section 2.4. 11.9.2 Compressibility Fluids are designated as compressible or incompressible depending on whether their density is variable or constant. We generally consider liquids to be incompressible and

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gases to be compressible and we will discuss certain exceptions to these ideas in subsequent chapters. The bulk modulus of elasticity, frequently referred to as just the bulk modulus, is a fluid property that characterizes the compressibility. It can be written as

=

dP dV/V

(11.24a)

dP d/

(11.24b)

or as

=− and has the dimensions N/m2 . 11.9.3 The Speed of Sound Disturbances that are introduced at some point in a fluid propagate at a finite velocity called acoustic velocity or speed of sound, c. It can be shown that the speed of sound is related to changes in pressure and density by  c=

dP d

1/2 (11.25a)

and use of Equation 11.24b gives us the alternate form   B 1/2 c= − 

(11.25b)

Moreover, for gases undergoing an isentropic process where P V k = C, a constant, we have  c=

kP 

1/2

or c = (k RT) 1/2

(11.25c)

where R in the SI System has units of J/kg-K. Often, in gas flows, even though the fluids are obviously compressible, density variations are negligibly small so that we may treat the flow as incompressible. A common criterion for such a consideration is the Mach number, which is defined as the ratio of the fluid ˆ to the speed of sound, c, in the fluid velocity of interest, V, M=

Vˆ c

(11.26)

If M < 0.2, the flow may be treated as incompressible with negligible error.

Example 11.4 A jet aircraft flies at an altitude of 9500 m where the air temperature is 226.41 K. Determine whether or not the compressible effects are significant when the aircraft speed is (a) 900 km/h and (b) 200 km/h.

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Solution Assumptions and Specifications (1) The flow is steady. (2) The air may be taken as an ideal gas. To determine the compressibility effects, we calculate the Mach number for the air at the two speeds. With k = 1.4, R = 287 J/kg-K (Table A.1 for air) and T = 226.41 K, Equation 11.25c gives c = (k RT) 1/2 = [(1.4)(287 J/kg-K)(226.41 K)]1/2 = 90,972 m2 /s2 = 301.61 m/s For Vˆ = 900 km/h (250 m/s), the Mach number is M=

Vˆ 250 m/s = = 0.829 c 301.61 m/s

Because M > 0.2, the compressibility effects are significant. For Vˆ = 200 km/h (55.5 m/s), the Mach number is M=

Vˆ 55.6 m/s = = 0.184 c 301.61 m/s

Because M < 0.2, the air flow over the aircraft may be regarded as incompressible with negligible error. 11.9.4 Surface Tension and Capillary Action We are familiar with the condition where a small amount of unconfined liquid forms a spherical drop. The reason for this occurrence is the attraction that exists between the liquid molecules. Within any particular drop, any molecule is surrounded by many others and the average force of attraction is the same in all directions. On the other hand, particles close to the surface will have an imbalance of net force because the number of adjacent particles are not uniform. The extreme condition, of course, is the density discontinuity. Particles right at the surface experience a relatively strong inward attractive force. With these ideas in mind, it is clear that some work must be done when a particle is moved toward the surface. If fluid is added to a drop, it will expand and additional surface will be created. The work needed to create new surface is the surface tension, designated by . Quantitatively, the surface tension is the work per unit area in N-m/m2 or force per unit length of interface in N/m. Both phases that are separated by an interface have the property of surface tension. The most common materials are water and air but many combinations are obviously possible. Surface tension, for a given interfacial composition, is a function of temperature and pressure although it is a much stronger function of temperature. Selected values for  for a few fluids in air are given in Table 11.1 and, for water in air at 1 atmosphere, the equation  = 0.1185(1 − 0.001325T)

N/m

(11.27)

can be used to provide  in N/m when T in K. The name, surface tension, suggests the presence of a skin-like membrane that holds the droplet together. Although this is not a correct physical concept quantitatively, the free surface can be treated as a membrane under uniform tension with magnitude, . Consistent

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TABLE 11.1

Surface Tensions of Some Fluids in Air at 1 atm at 20◦ C as a Function of Temperature (Extracted from Handbook of Chemistry and Physics) σ, N/m

Fluid Ammonia Ethyl alcohol Gasoline Glycerin Kerosene Mercury Soap solution SAE 30 oil

0.021 0.028 0.022 0.063 0.028 0.440 0.025 0.035

with this concept of surface tension, is the idea that the curved surface will be associated with a pressure difference. Because the sphere represents the minimum surface area for a prescribed volume, we show a free-body diagram of the forces in a hemisphere of radius, r (Figure 11.17). The difference in pressure between the inside and outside of the hemisphere, P, acting over the area, r 2 , is balanced on the hemisphere periphery by the force due to the surface tension, 2 r . Hence, the required force balance is

r 2 P = 2 r  which shows that the difference in pressure is P =

2 r

(11.28)

For a soap bubble, which has an extremely thin wall, the pressure difference across the two interfaces present will be P =

4 r

(11.29)

We note that as the radius of curvature, r , increases, the resulting P will decrease. Thus, in the case of no curvature (a flat surface), there is no net pressure difference due to surface tension. An important result of the surface-tension-caused pressure difference across a curved surface is the phenomenon of capillary action. This effect, which can cause a liquid column to rise in a tube, is a function of how well a liquid wets a solid boundary. The contact angle 2πrσ

πr2ΔP

FIGURE 11.17 Free-body diagram of a hemispherical droplet showing the balance between pressure and tension forces.

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Introduction to Thermal and Fluid Engineering Gas Liquid θ

θ

FIGURE 11.18 The contact angle associated with a gas-liquid-vapor interface for the nonwetting case.

associated with a gas-liquid-solid interface is the indicator of a wetting or nonwetting case. As shown in Figure 11.18, the contact angle, , is measured in the liquid and the delineation between wetting and nonwetting liquids is in the magnitude of the contact angle. For wetting liquids,  < 90◦ and for nonwetting liquids,  > 90◦ . For example, mercury in contact with a clean glass tube has a value of  ≈ 130◦ . For water in a clean glass tube,  may be taken as  = 0◦ because the water almost completely wets the surface. In Figure 11.19a, we see that a small diameter tube has been inserted into a pool of water. The liquid level interface in the tube makes an angle, , with the wall of the tube and rises a distance, h, above the level of the water in the pool. In this case, an attraction or adhesion between the wall of the tube and the liquid molecules overcomes the mutual attraction of the molecules (cohesion) at the surface of the liquid. Here, the water appears to be “pulled up” into the tube and is said to wet the surface of the tube. Figure 11.19b shows a tube inserted into a pool of mercury where the adhesion of the mercury molecules to the solid surface is weak in comparison with the cohesive forces at the liquid surface. In this case, the mercury is depressed a distance, h, below the level of the pool of mercury. A free-body diagram of a wetting liquid within a tube is shown in Figure 11.19c. Here the vertical force due to the surface tension 2 r  cos  must balance the force due to the weight of the fluid volume, V = r 2 h  r 2 hg 2πrδ

h h

(a)

(b)

πr2h (c)

FIGURE 11.19 (a) A tube immersed in water, (b) a tube immersed in mercury, and (c) a free-body diagram of a wetting liquid inside a tube.

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Hence, 2 r  cos  =  r 2 hg and the height, h, is given by h=

2 cos  gr

(11.30)

Example 11.5 A 5-mm-radius glass tube is inserted into a pool of mercury at 20◦ C. The surface tension of the mercury and air is 0.47 N/m and the contact angle is 130◦ . Determine the distance that the mercury is depressed below the pool level (Figure 11.19b).

Solution Assumptions and Specifications (1) Steady state exists. (2) The acceleration of gravity is assumed to be 9.81 m/s2 . (3) The density of water is 1000 kg/m3 . (4) Mercury is an incompressible liquid and has a specific gravity of 13.6. For mercury,  = 13,600 kg/m3 and use of Equation 11.30 gives h=

2 cos  2(0.47 N/m)(cos 130◦ ) = gr (13,600 kg/m3 )(9.81 m/s2 )(5 × 10−3 m)

or h = −9.06 × 10−4 m

11.10

(0.906 mm) ⇐

Summary

In this introductory chapter to the subject of fluid mechanics, a number of concepts and definitions have been introduced. Most of them will be encountered quite often in the remainder of this book. The concept of a fluid being a material that deforms continuously when subjected to a shearing stress led to a discussion of stress and how to characterize stress at a point, as well as the manner in how it changes throughout a fluid continuum. A property that can vary from point to point was examined, and the rate of change was seen to vary from zero along an isoline to a maximum in the direction of the gradient. In this book, gases and liquids, both fluids, are considered as being continuously distributed throughout the region so that the fluid may be considered as a continuum. The density, , was the primary property employed to determine if the continuum concept is appropriate. Laminar and turbulent flow were discussed. Smooth and orderly flow is termed laminar. The flow is layer-like and fluid particles in a given layer retain their identity as long as laminar flow continues. In contrast, turbulent flow is characterized by apparently random motion of large numbers of fluid particles.

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The fluid property viscosity was defined and its role in producing a force between a moving surface and a viscous fluid was illustrated. Newton’s viscosity relationship of the form  yx =

d Vˆ x dy

is the defining relationship for viscosity.

11.11

Problems

Evaluation of the Gradient 11.1: A scalar field, , is represented by  = xy2 (z − 1) 3 Determine grad  and evaluate grad  at x = 1, y = 3, and z = −1. 11.2: A scalar field, , is represented by  = x 3 ( y − 2) Determine grad  and evaluate grad  at x = 2, y = 2, and z = 3 11.3: A scalar field, , is represented by  = (x − 2) 2 yz2 Determine grad  and evaluate grad  at x = −1, y = −2, and z = 3. 11.4: A scalar field, , is represented by  = x 2 y3 z4 Determine grad  and evaluate grad  at x = 1, y = −2, and z = −3. 11.5: A scalar field, , is represented by  = x(x 2 − 1) yz Determine grad  and evaluate grad  at x = 1, y = 2, and z = −3. 11.6: A scalar field, , is represented by  = xy( y + 3)z2 Determine grad  and evaluate grad  at x = 1, y = 3, and z = 5. 11.7: A scalar field, , is represented by  = x 2 (z2 − 4z) Determine grad  and evaluate grad  at x = −1, y = 3, and z = −5. 11.8: A scalar field, , is represented by  = 3x 2 y + 4y2 Determine grad  at the point (3, 5).

(11.19)

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11.9: Show that the unit vectors r and θ are related to the unit vectors x and y by r = x cos  + y sin  and θ = x sin  + y cos  11.10: Using the results of Problem 11.9, show that dr =θ d

dθ = −r d

and

11.11: Using the results of Problem 11.9, and, in addition ∂ ∂x

= cos 

∂ ∂r

+

cos  ∂ r ∂

−

sin  ∂ r ∂

and ∂ ∂y

= sin 

∂ ∂r

transform the operator ∇ to the cylindrical coordinates, r, , and z. 11.12: Using the expression for the gradient in polar coordinates, determine the gradient of (r, ) and the point of maximum gradient when A and a are constants and 

a2  = Ar sin  1 − 2 r



11.13: A pressure field is given by    x 2 Vˆ t 1 ˆ2 xyx ∞ P = Po + V ∞ 2 +3 + 2 L3 L L where x, y, and z are space coordinates, t is time, and Po , , Vˆ ∞ , and L are constants. Find the pressure gradient. Density 11.14: Which of the following conditions are flow properties and which are fluid properties? pressure temperature velocity density

stress

specific heat

pressure gradient

speed of sound

11.15: The molecular weight of carbon monoxide is 28.01. Determine its density in kg/m3 when the absolute pressure is 310 kPa and the temperature is 50◦ C. 11.16: The molecular weight of chlorine is 70.9. Determine its density in kg/m3 at a gage pressure of 101 kPa and a temperature of 27◦ C. 11.17: A tire having a volume of 0.08m3 contains air at a gage pressure of 210 kPa and a temperature of 20◦ C. Determine (a) the density of the air and (b) the weight of the air continued in the tire. 11.18: A compressed air tank contains 10 kg of air at a temperature of 80◦ C. If a gage on the tank reads 250 kPa, determine the volume of the tank.

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11.19: A large dirigible having a volume of 100,000 m3 contains helium under standard atmospheric conditions (101 kPa and 15◦ C). Determine the density and the total weight of the helium. 11.20: A quart of SAE 30 oil weighs about 0.84 kg. Assume that the density of water is 1000 kg/m3 and determine the mass density, the specific weight, and the specific gravity of the oil. 11.21: Gasoline has an approximate specific weight of 7150 N/m3 . Assume that the density of water is 1000 kg/m3 and determine its mass density, specific volume, and specific gravity. 11.22: A conical tank has a base radius of 25 cm and a height of 50 cm. Assume that the density of water is 1000 kg/m3 , and determine the water level above the base of the tank if 0.024 m3 of water are poured into the tank. 11.23: Under standard conditions, a quantity of gas weighs 0.15 N/m3 . Determine its density and specific volume. 11.24: Determine the specific weight of a gas having a specific volume of 0.125 m3 /kg. 11.25: A gas begins to deviate from the continuum when it contains less than about 1012 molecules per cubic millimeter. To what absolute pressure in pascals at 300 K does this limit for air correspond? 11.26: For a fluid density, , in which solid particles of density, s , are uniformly dispersed, show that if x is the mass fraction of solid in the mixture, the density is given by s  mixture = x + s (1 − x) 11.27: Assuming that the fluid of density, , in Problem 11.26 obeys the ideal gas law, obtain the equation of state of the mixture, that is, P = f (s , RT/M, mix , x). Viscosity 11.28: Crude oil having a viscosity of 0.046 kg/m-s is contained between parallel plates. The bottom plate is fixed and the upper plate moves when a force is applied (Figure P11.28). If the distance between the two plates is 5 mm and the effective area of the upper plate is 0.096 m2 , determine the value of P required to move the upper plate with a velocity of 6 cm/s. 0.096 m2

5 mm

Oil, μ = 0.046 kg/m-s

P, V = 6 m/s

Fixed Plate FIGURE P11.28

11.29: A cubical block 0.250 m on a side and having a mass of 0.2548 kg slides down a 22.5◦ incline on a film of oil that is 0.626- m thick. Assume a linear velocity profile in the oil having a viscosity of 0.007 kg/m-s and determine the velocity of the block. 11.30: A Newtonian fluid having a density of 920 kg/m3 and a kinematic viscosity of 5 × 10−4 m/s2 flows past a fixed surface. The velocity profile near the surface is shown in Figure P11.30. Determine the magnitude and the direction of the shearing stress

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ˆ and  in units of m/s developed on the plate expressing your result in terms of V and m, respectively.  V y V x

δ

x V V

=

3 y 1 4 – 2 δ 2 6

3

x FIGURE P11.30

11.31: A steel (7850 kg/m3 ) shaft that is 6 cm in diameter and 30-cm long falls of its own weight inside a vertical open tube that is 4.03 cm in diameter. The clearance, assumed to be uniform, contains a film of glycerin at 20◦ C. Determine the velocity of the shaft at terminal conditions. 11.32: In Figure P11.32, a solid cone of angle 2, base radius, ro , and density c is rotating with angular velocity o inside a conical seat. The clear space, designated by h, is filled with an oil of viscosity, . Neglecting air drag, derive an expression for the time required to reduce the angular velocity of the cone from o to 0.10 o .

ω(t)

Base Radius r0

Oil

2θ h

FIGURE P11.32

11.33: A disk of radius R rotates at an angular velocity, , inside an oil bath of viscosity as indicated in Figure P11.33. Assuming a linear velocity profile and neglecting shear on the outer disk edges, derive an expression for the viscous torque on the disk. ω

Clearance h

Oil R

R FIGURE P11.33

11.34: A 35-mm-diameter shaft is pulled through a cylindrical bearing is shown in Figure P11.34. The lubricant that fills the 0.3-mm gap between the shaft and the bearing

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Introduction to Thermal and Fluid Engineering is an oil having a kinematic viscosity of 8 × 10−4 m/s2 and a density of 910 kg/m3 . Assume the velocity distribution in the gap to be linear and determine the force, P, required to pull the shaft at a velocity of 4 m/s. Lubricant Bearing

P

0.5 m FIGURE P11.34

11.35: Air at 20◦ C forms a boundary near a solid wall (Figure P11.35) with a sine-waveshaped velocity profile. The boundary layer thickness is 6 mm and the peak velocity is 10 m/s. Compute the shear stress in the boundary layer in pascals at y equal to (a) 0 mm, (b) 3 mm, and (c) 6 mm. V max = 10 m/s Peak 6 mm y FIGURE P11.35

11.36: A layer of water flows down an inclined fixed surface with the velocity profile shown in Figure P11.36. Determine the magnitude and direction of the shearing stress that ˆ = 2 m/s and h = 0.10 m. the water exerts on the fixed surface for V

V

h y

y y3 =2 – 2 h h  V

x V

V x

FIGURE P11.36

11.37: A piston that is 75-cm long moves through a 12.5-cm-diameter cylinder. The film separating the piston and the cylinder is 0.025-cm thick and contains oil with a viscosity of 0.975 kg/m-s. Determine the force required to maintain a velocity of 6 m/s. 11.38: A piston having a mass of 0.75 kg, a length of 15 cm and a diameter of 12.5 cm slides through a 12.005-cm inner-diameter vertical pipe. The piston is decelerating at a rate

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357

of 0.725 m/s2 when the velocity is 7.25 m/s. Determine the viscosity of the oil that occupies the space between the piston and the cylinder. 11.39: Two 15-cm-long cylinders, of diameters 7.5 cm and 8.0 cm, fit inside each other. The gap is filled with an oil having a viscosity of 0.425 kg/m-s. Assuming the outer cylinder to be fixed and that the velocity profile in the oil is linear, determine (a) the torque and (b) the power required to maintain a rotation of the inner cylinder at 6.25 rad/s. 11.40: A 25-cm-diameter shaft 75 cm in length is pulled through a cylindrical bearing. Oil having a kinematic viscosity of 8.75 × 10−4 m/s2 and a specific gravity of 0.92 fills the 0.325-mm gap between the stationary and moving surfaces. Assuming that the velocity distribution in the gap is linear, determine the force required to pull the shaft through the bearing at a velocity of 4 m/s. 11.41: A disk of radius, r , rotates with an angular velocity of (Figure P11.41). The clearance, as indicated in the figure is h = 0.55 mm, the disk radius is r = 0.80 mm, the angular velocity is = 0.375 rad/s and the viscosity of the oil is = 0.008 kg/m-s. Neglecting end effects and assuming the velocity profile in the fluid to be linear, determine the damping torque. ω = 0.375 rad/s

h = 0.55 mm FIGURE P11.41

11.42: A 25-cm-diameter plunger slides inside a 25.02-cm-diameter cylinder. The annular space is filled with oil having a kinematic viscosity of 3.75 × 10−4 m/s2 and a specific gravity of 0.865. Assume a linear velocity distribution in the oil and, when the plunger moves at 0.18 m/s, determine the frictional resistance when 2.75 cm of the plunger is in the cylinder. 11.43: A 15-cm-diameter shaft is turning in a 15.05-cm-diameter sleeve that is 20-cm long. The gap is filled with an oil having a viscosity of 0.0875 kg/m-s. If the velocity distribution in the oil is linear and the shaft turns at 90 rpm, determine the rate of heat generation. 11.44: A 40-cm square block slides on oil with viscosity of 0.81 kg/m-s over a large smooth surface. The film between the block and the surface is 5-mm thick and the velocity distribution in the oil is linear. Determine the force required to move the block at a velocity of 4 m/s. 11.45: Two plates are separated by an 8-mm gap that contains oil with a viscosity of 0.450 kg/m-s. The bottom plate is stationary and the upper plate moves at 3.75 m/s; the velocity distribution in the oil is linear determine the shear stress. 11.46: A large plate moves with a velocity of 5 m/s over a stationary plate on a layer of oil ( = 0.42 kg/m-s) that is 5 mm thick. A parabolic profile is assumed for the velocity distribution. Determine the shear stress on the upper plate. 11.47: A 24-kg cube slides down a 20◦ inclined plane on a 4-mm-thick oil film. The contact area between the cube and the plane is 0.25 m2 and the velocity profile may be assumed linear; the velocity is 6 m/s. Determine the viscosity of the oil. 11.48: A shaft that is 8 cm in diameter is rotated at 2400 rpm inside a sleeve that is 8.2 cm in diameter. The length of the sleeve is 20 cm and oil, with a kinematic viscosity of

358

Introduction to Thermal and Fluid Engineering 48×10−4 m/s2 and a specific gravity of 0.925, fills the gap. If the velocity distribution in the oil is linear. Determine the power required to rotate the shaft.

11.49: Two large plane surfaces are separated by a gap of 1.2 cm, which is filled with oil having a viscosity of 0.80 kg/m-s. A thin plate having an area of 0.50 m2 and a thickness of 2 mm is placed midway between the two large plates and dragged through the gap at 0.175 m/s. If the velocity distribution in the oil is linear, determine the force required. Additional Fluid Properties 11.50: A vessel that contains 1.25 m3 of water at atmospheric pressure is heated from 60 to 95◦ C. How much water must be removed to keep the volume unchanged? 11.51: A vertical cylindrical tank having a base diameter of 10 m and a height of 6 m is filled to the top with water at 20◦ C. How much water will overflow if the water is heated to 60◦ C? 11.52: A liquid in a cylinder has a volume of 1200 cm3 at 1.25 MPa and a volume of 1188 m3 at 2.50 MPa. Determine its bulk modulus of elasticity. 11.53: A differential pressure of 10 MPa applied to 0.25 m3 of a liquid causes a volume reduction of 0.005 m3 . Determine the bulk modulus of elasticity. 11.54: The bulk modulus of elasticity for water is 2.205 GPa. Determine the change in pressure required to reduce a given volume by 0.75%. 11.55: Given the bulk modulus of elasticity for water is 2.205 GPa, determine the change in volume of 1 m3 of water when subjected to a change in pressure of 20 MPa. 11.56: Water in a container is originally at 100 kPa. The bulk modulus of elasticity for water is 2.205 GPa and the water is subjected to a pressure of 120 MPa. Determine the percentage decrease of its volume. 11.57: Given the bulk modulus of elasticity for water of 2.205 GPa, determine the change in pressure required to reduce a given volume by 1.25%. 11.58: The viscosity of water is 0.0100 poise at 20◦ C. Determine its kinematic viscosity. Surface Tension and Capillary Action 11.59: A small droplet of water at 25◦ C is in contact with air and has a diameter of 0.05 cm. The difference in pressure between the inside and outside of the droplet is 565 Pa. Determine the value of the surface tension. 11.60: Determine the height to which water at 68◦ C will rise in a clean capillary tube having a diameter of 0.2875 cm. 11.61: Two clean and parallel glass plates, separated by a gap of 1.625 mm, are dipped in water. If  = 0.0735 N/m, determine how high the water will rise. 11.62: A glass tube having an inside diameter of 0.25 mm and an outside diameter of 0.35 mm is inserted into a pool of mercury at 20◦ C such that the contact angle is 130◦ . Determine the upward force on the glass. 11.63: In Problem 11.62, determine the depth of the depression. 11.64: Determine the capillary rise for a water-air-glass interface at 40◦ C in a clean glass tube of radius, r = 1 mm. 11.65: Determine the capillary rise if the tube in Problem 11.64 is immersed in mercury at 20◦ C such that the contact angle is 126◦ .

Introduction to Fluid Mechanics

359

11.66: Determine the difference in pressure between the inside and outside of a soap film bubble at 20◦ C if the diameter of the bubble is 4 mm. 11.67: At 25◦ C, determine the maximum diameter clean glass tube necessary to keep the capillary height at a maximum of 2.5 mm. 11.68: Assuming that the difference in the inside and outside radii of a soap bubble is negligible, determine the surface tension for a 2.25-cm-diameter bubble subjected to a difference in pressure between the inside and outside of P = 31 Pa. 11.69: Two vertical, parallel, clean, glass plates are spaced a distance of 3 mm apart and they are placed in water. Determine how high the water will rise between the plates due to capillary action. 11.70: An open, clean glass tube, having a diameter of 3 mm, is inserted vertically into a dish of mercury at 20◦ C. Determine how far the column of mercury in the tube will be depressed. 11.71: At 60◦ C, the surface tensions of water and mercury are water:  = 0.0662 N/m and mercury:  = 0.47 N/m. Determine the capillary height changes in these two fluids when they are in contact with air in a glass tube of diameter 0.55 mm. Contact angles are 0◦ for water and 130◦ for mercury. 11.72: Determine the diameter of the glass tube necessary to keep the capillary-height change of water at 30◦ C less than 1 mm.

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12 Fluid Statics

Chapter Objectives •

To develop the basic equation of fluid statics.

•

To describe the use of manometers to measure pressure.

To evaluate hydrostatic forces on planar and nonplanar surfaces immersed in a fluid. • To develop the concept of buoyancy. •

•

To examine the stability of floating objects.

•

To describe hydrostatic forces under conditions of uniform rectilinear acceleration.

12.1

Introduction

In Chapter 11, we defined a fluid according to its response to a shearing stress and we commented that a fluid at rest would therefore not be subjected to shear stresses. In this chapter we will consider the behavior of fluids at rest, that is, in a static condition. We should distinguish, at the outset, between inertial and noninertial reference frames. In general, we will consider a static situation to be one that is stationary relative to the earth’s surface; this is what is meant by an inertial reference frame. If a fluid is stationary relative to a coordinate system that has some acceleration of its own, such a reference frame is termed noninertial. An example of a noninertial reference frame would be a fluid inside an aircraft as it executes a maneuver. Our considerations will be, in general, for inertial reference frames and we will note, as exceptions, those occasions where noninertial reference frames are employed.

12.2

Pressure Variation in a Static Field

In the absence of shear stresses a fluid will experience only gravitational and pressure forces. A representative fluid element of differential size in a static fluid is shown in Figure 12.1. Its dimensions, as indicated, are d x, dy, and dz. Fluid pressure, P(x, y, z), which acts on all six faces of the element, will produce force components in each of the coordinate directions. Because pressure always acts inward, the x-component pressure forces, which result from the pressure exerted on the x-faces are 361

362

Introduction to Thermal and Fluid Engineering P+

∂p dz dxdy ∂z

Pdydz

dz P + ∂P dy dxdz ∂y

Pdxdz dx dy ∂P P+ dz dydz ∂x Pdxdy z

x y

FIGURE 12.1 Pressure forces exerted on a static fluid element.

expressed as 

 Pdyd x − P +

∂P ∂x





d x d yd x x

After performing the cancellation, we write the net x-component pressure force: Px = −

∂P ∂x

d xd ydz

Doing the same operation in the y-coordinate direction, we have     ∂P Pdzd x − P + dy dzd x y ∂y which reduces to Py = −

∂P ∂y

d xd ydz

and in the z-coordinate direction     ∂P Pd xdy − P + dz d xd y z ∂z which becomes the z-component Pz = −

∂P ∂z

d xd ydz

The gravitational force is expressed, in general form, as g d xd ydz where the direction of g is left in a completely general form.

Fluid Statics Now, because

363 

F = 0, we have  F = Px x + Py y + Pz z + g d xd yd x = 0

or  −

∂P ∂x

x+

∂P ∂y

y+

∂P ∂z

 z d xd ydz + g d xd yd x = 0

This expression can be simplified further to the form ∂P ∂z

x+

∂P ∂y

y+

∂P ∂z

z = g

(12.1)

Recalling the definition of the gradient from Chapter 11, Equation 12.1 may be written in the more compact form ∇ P = g

(12.2)

Equation 12.2 is generally referred to as the basic equation of fluid statics. We will devote the remaining sections of this chapter to applications of this fundamental relationship. As a final comment before proceeding with applications, we observe that Equation 12.2 is a general vector relationship that is valid for any orthogonal coordinate system even though it was derived from consideration of an element in Cartesian coordinates. For any element, which is essentially a point, an orthogonal coordinate system would consist of three unit vectors that are mutually perpendicular. We could designate these directions as x, y, and z or r, , and z or in any way we choose. The choice would depend on how we wish to build up (or integrate) from the point to conform to boundaries or coordinates conveniently.

12.3

Hydrostatic Pressure

We have already discussed the consequence of Equation 12.2 for the directions of maximum pressure variation and of zero (constant pressure) variation. A direct application of this idea involves a pressure measuring device called a manometer. We can use a U-tube manometer (Figure 12.2) to evaluate the pressure at point A in the chamber by measuring the levels of the manometer fluid on the two sides of the U-tube. If we start with Equation 12.2, we may write for Cartesian coordinates ∂P ∂x

x+

∂P ∂y

y+

∂P ∂z

z = −gz

The next step is to equate the coefficients of the unit vectors and, in doing so, we obtain two scalar equations ∂P ∂x ∂P ∂y

=0

(12.3a)

=0

(12.3b)

364

Introduction to Thermal and Fluid Engineering Patm

z2 g

A z1 B

C

FIGURE 12.2 A U-tube manometer. Note that g is in the negative z direction.

showing that the pressure does not vary in the x and y directions, and a third: ∂P ∂z

= −g = −

(12.3c)

Each of these scalar relationships provides us with some information and the useful one is Equation 12.3c. If we apply this equation to our manometer between points C and D, we may separate the variables d P = −g dz and then integrate between the appropriate limits 

PD

 d P = −g

PC

z2

dz 0

to obtain PC − PD = gz2 = z2

(12.4)

or, because PD is atmospheric pressure PC = Patm + gz2 = Patm + z2

(12.5)

This result indicates that, for a vertical tube open to the atmosphere, the pressure at a point below the free liquid surface is greater than atmospheric by the amount gz, which is clearly proportional to, z, the height of fluid. This expression will be used without detailed derivation in future discussion in examples involving manometers. In the development of Equation 12.5, the fluid density was taken as constant. In one of the examples soon to follow, we will illustrate the procedure for variable density. Because points B and C are at the same level and are connected by a continuous column of manometer fluid, our earlier discussion indicates that they are at the same pressure. The pressure at point A can be evaluated, beginning with the statement of this equality PB = PC

Fluid Statics

365

by using s for the “system” fluid and m for the “manometer” fluid and then applying Equation 12.4 twice PB = PA + s gz1 and PC = Patm + m gz2 Substitution then yields the result PA + s gz1 = Patm + m gz2 which may be written as PA = Patm + m gz2 − s gz1

(12.6)

PA, g = m gz2 − s gz1

(12.7)

or

Equation 12.5 indicates that the absolute pressure at point C is the sum of the atmospheric pressure and the quantity gz2 . The difference between atmospheric pressure and absolute pressure is designated the gage pressure if gz2 is positive or the vacuum pressure if gz2 is negative. An alternative way of writing Equation 12.4 in terms of the gage pressure, in this case, PC, g , is PC, g = gz2

(12.8)

Examples 12.1 and 12.2 that now follow will illustrate the application of Equation 12.5 to configurations that involve manometers.

Example 12.1 A fan draws air from the atmosphere through a round duct as shown in Figure 12.3. A differential manometer connected to an opening in the duct wall shows a vacuum pressure of 2.5 cm of water. Determine the air pressure at this point.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The water in the manometer is static and incompressible.

Airflow

Fan

A z1 z2 = 2.5 cm B

C Water (density = ρw)

FIGURE 12.3 Airflow through a duct for Example 12.1.

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Introduction to Thermal and Fluid Engineering

(3) The acceleration of gravity is assumed to be 9.81 m/s2 . (4) The density of the water is specified. We assume that the temperature at point A is at 15◦ C where a = 1.234 kg/m3 and we note at the outset that we already know the pressure at point A, PA = 2.5 cm of water vacuum. Thus, PC = Patm = PB and with the densities of air and water designated with subscripts a and w PB = PA + a gz1 + w gz2 or PA = Patm − a gz1 − w gz2 The gage pressure at point A will be PA, g = −a gz1 − w gz2 and this can be written as



PA, g

a = w g − z1 − z2 w



The density ratio at the standard conditions with T = 15◦ C can be evaluated. For w = 1000 kg/m3 as a 1.234 kg/m3 = = 1.234 × 10−3 3 w 1000 kg/m and it is clear that the first term within the parentheses will be negligibly small and may thus be neglected. This leaves PA, g = −w gz2 which may be converted to Pascals in the usual process of unit conversion PA, g = −

(1000 kg/m3 )(9.81 m/s2 )(0.025 m) 1 kg-m/N-s2

= −245.3 Pa ⇐

Example 12.2 Determine the gage pressure in the bulb at point Afor the conditions shown in Figure 12.4.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluids are static and incompressible. The three material densities may be designated with the subscripts a, oil, and Hg for the air, oil, and mercury, respectively. We next apply the principles that have been developed

Fluid Statics

367 Air

B

Oil, SG = 0.82

C 20 cm

40 cm

40 cm 12 cm E

D A

Mercury

Water FIGURE 12.4 The compound manometer discussed in Example 12.2.

to express the following: PD = PE = Patm + oil g(40 cm) PD = PC + a g(20 cm) + Hg g(12 cm) PB = PC PA = PB + w g(40 cm) An algebraic exercise that combines these expressions leads to the expression for PA PA = Patm + w g(40 cm) + oil g(40 cm) − a g(20 cm) − Hg g(12 cm) If all of the densities are expressed in terms of w we obtain PA = Patm + w g(40 cm) + 0.82w (40 cm) a − w g(20 cm) − 13.6w g(12 cm) w which can be simplified, with w = 1000 kg/m3 , to   a PA = Patm + w g 40 cm + 32.8 cm − (20 cm) − 163.2 cm w As before, because a /w  1, we can neglect the air pressure term. Combination of the rest of the terms yields the gage pressure PA, g = PA − Patm = w g(−90.4 cm) or PA, g = 8.87 kPa (vacuum) ⇐ The next example considers the density of a gas that varies with altitude.

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Introduction to Thermal and Fluid Engineering

Example 12.3 A temperature of 5◦ C and a pressure of 0.612 bar are noted on the instrument panel of an aircraft in level flight at an altitude of h m. Surface conditions are 760 mm of mercury and 25◦ C. Assuming that the atmosphere behaves as a perfect gas and that the air temperature variation between the surface of the earth and the aircraft is linear, estimate the altitude of the aircraft.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The standard barometer is 760 mm of mercury. (3) Atmospheric air may be considered as a perfect gas. (4) The air temperature variation is linear with altitude. (5) The acceleration of gravity has a value of g = 9.81 m/s2 . At the surface of the earth, standard atmospheric pressure of 760 mm of mercury is specified and, in accordance with the problem statement, the specified linear temperature variation would be described by T = To −

T z h

where To = 25◦ C

and

T = 25◦ C − 5◦ C = 20◦ C

(20 K)

Moreover, we have the equation of state for the ideal gas =

P RT

where, by Table A.1 in Appendix A, R = 287 N-m/kg-K. The basic equation of fluid statics gives dP P = −g = − g dz RT We next relate the temperature variation with altitude and separate the variables to obtain ⎛ ⎞ dP ⎜ = −⎝ P

g/R ⎟ dz T ⎠ To − z h

Now, integration gives 

P



h

dz T Po 0 To − z h   P g h T ln = ln 1 − Po R T h

or

dP g =− P R

  P T gh/RT = 1− Po To

Fluid Statics

369

Here we note that To need not be converted to K as it plays no thermodynamic role in the foregoing relationship. Then we evaluate P/Po with Po = 1.013 × 105 Pa (760-mm Hg) as P 460-mm Hg = = 0.6053 Po 760-mm Hg and gh (9.81 m/s2 )h = = (1.709 × 10−3 m−1 )h RT (287 N-m/kg-K)(20 K) so that

  P T gh/RT = 1− Po To −1 −3   20◦ C (1.709×10 m )h 0.6053 = 1 − ◦ 25 C −1 −3 = (0.20) (1.709×10 m )h

(1.709 × 10−3 m−1 )h = 0.312 h = 182.5 m ⇐

12.4

Hydrostatic Forces on Plane Surfaces

When a solid object is immersed in a fluid, the force exerted on each of its surfaces is due to fluid pressure. Accordingly, we may use our fluid statics relationships to evaluate these forces. A plane surface immersed in a liquid at an angle  relative to the liquid surface is shown in Figure 12.5. The total surface area is, S, the fluid has a uniform density, , and we are using h as the depth and y as the length along the surface as shown. For each point on the surface, S, at the same depth, h, the gage pressure will be gh. This will be the case for the differential area, dS, which is at a uniform depth, h = y sin , below the fluid surface. The pressure force on dS is dF = PdS = gy sin dS and the total force can be obtained by integration   F = dF = g sin  y dS

(12.9)

(12.10)

S

If we recall the definition of the centroid of an area  1 y¯ = y dS S

(12.11)

we may write Equation 12.10 in an alternate form as F = g sin S y¯

(12.12)

Equation 12.12 indicates that the force on a submerged surface is equal to the pressure evaluated at its centroid multiplied by the submerged surface area, S.

370

Introduction to Thermal and Fluid Engineering Liquid Surface h

_

hcp

θ y

h _

y ycp

dS

FIGURE 12.5 A submerged plane surface.

If we were to consider a pressure force on a given surface in a static analysis, we would need both its magnitude, which has just been evaluated, and its location, which has yet to be established. The location of interest is designated as the center of pressure and its position will be evaluated by referring, once again, to Figure 12.5. We seek the value of y = ycp , where the total pressure force will act so as to produce the same moment as the distributed pressure. Using point 0 as reference, we have   F ycp = y dF = yP dS (12.13) S

S

and substituting for the pressure, we have  F ycp = g sin y2 dS

(12.14)

S

Using Equation 12.12 for the force provides  Icp 1 ycp = y2 dS = S y¯ S S y¯

(12.15)

where Icp is the moment of inertia of the surface area about its center of pressure. An alternative expression can be obtained if the moment of inertia is evaluated about an axis through its centroid rather than the center of pressure. The transfer theorem relating Io (the moment of inertia about the centroidal axis of S) and Icp is Icp = Io + S y¯ 2 Substituting this expression into Equation 12.15, we obtain ycp =

Io + S y¯ 2 Io = + y¯ S y¯ S y¯

(12.16)

Fluid Statics

371

h/2 x

x

A = bh Ixx = bh3/12

h/2 b

b x

x a

R x

A = πab Ixx = πab3/4

x A = πR2 Ixx = πR4/4

x

x a

2h/3 x

x

b

4b 3π A = πab/2 Ixx = 0.109757ab3

h/3 b/2

b/2

x

A = bh/2 Ixx = bh3/36

x R

4R 3π

A = πR2/2 Ixx = 0.109757R4

FIGURE 12.6 Moments of inertia and centroidal axes for six plane areas.

which indicates that the center of pressure will always be below the centroidal axis of the area by the amount Io /S y¯ . Figure 12.6 provides a guide to the location of the centroidal axes and the equations for the moments of inertia about the centroidal axes for several plane areas.

Example 12.4 Determine the minimum fluid level h for which the rectangular gate shown in Figure 12.7a will rotate in a counterclockwise direction. The cross section of the gate (Figure 12.7b) is 2 m ×2 m and friction in the hinge may be neglected.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is no flow of fluid. (3) The fluid is incompressible. (4) The friction in the hinge may be neglected.

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Introduction to Thermal and Fluid Engineering

Pg = 6 × 104 Pa h

2m

A

2m

A

Section A-A (b)

(a)

FIGURE 12.7 A rectangular gate separating the fluid column and air chamber in Example 12.4.

(5) The acceleration of gravity may be taken as g = 9.81 m/s2 . (6) The density of water may be taken as 1000 kg/m3 . Figure 12.8 is a free-body diagram of the gate showing the water force Fw . With y = (h − 1) m, w = 1000 kg/m3 and  = 90◦ Fw = w gS y¯ = (1000 kg/m3 )(9.81 m/s2 )(2 m)(2 m)[(h − 1) m] = 39, 240(h − 1) N The location of this force is at ycp from the hinge ycp = 1 m +

Io S y¯

and, evaluating Io as Io =

bh 3 (2 m)(2 m) 3 4 = = m4 12 12 3 Point 0

y– = (h – 1) m h y0 Fw lm

FIGURE 12.8 A detail of the gate in Example 12.4 showing the water force.

Fluid Statics

373

h–2 2m

y Fw

y

dy 2m

Fa h

ds

FIGURE 12.9 Detail of gate in Example 12.4 showing both the water force and the air force.

we have ycp = 1 m + = 1m +

4/3 m4 (2 m)(2 m)[(h − 1) m] 1m (3h − 2) m = 3(h − 1) 3(h − 1)

We may proceed to solve for h by taking moments about the hinge. Fw ycp = Pa S y¯ cp (39,240 N)(h − 1)

3h − 2 = (6 × 104 N/m2 )(4 m2 )(1 m) 3(h − 1)

3h − 2 m =

24 × 104 N-m 13,080N

3h − 2 m = 18.35 m 3h = 20.35 m h = 6.78 m ⇐ We can check this result via an alternative and equivalent procedure. The net force on dS in Figure 12.9 is dF = ( Pw − Pa )dS = ( Pw − Pa )2 dy With Pw and Pa given by Pw = Patm + w g(h − 2 + y) and Pa = Patm + Pg we have dF = [Patm + w g(h − 2 + y) − ( Patm + Pg )]2 dy

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Introduction to Thermal and Fluid Engineering

or dF = [w g(h − 2 + y) − Pg ]2 dy The problem statement requires that the following relation holds:   2m dm = y dF = 0 A

0

Thus, 

2m

y[w g(h − 2 + y) − Pg ]2 dy = 0

0



2m

0

 w g

[w g(hy − 2y + y2 ) − yPg ]dy = 0

  hy2 y3 2 m y2 2 m 2 −y + − Pg  = 0 2 3 0 2 0   8 w g 2h − 4 m + m = 2Pg 3

Then 2h −

2Pg 4 m= 3 w g h=

Pg 2 m+ 3 w g

=

2 6 × 104 N/m2 m+ 3 (1000 kg/m3 )(9.81 m/s2 )

=

2 m + 6.11 m = 6.78 m ⇐ 3

We now turn to Design Example 5, which deals with the significant dimension in a retaining wall.

12.5

Design Example 5

An earth-filled retaining wall having a density of 1700 kg/m3 and with the profile shown in Figure 12.10a is used to hold back seawater with a density of 1025 kg/m3 . Determine the dimension, L , which will produce a zero moment about point O under the foregoing conditions.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The densities of the retaining wall and seawater are specified. (3) All forces and moments are evaluated per meter of retaining wall depth.

Fluid Statics

375 L

L

B

B

2

H 3

Fy

7.5 m

Earth Fill

7.5 m Fx H 3

60° A

60° A

0

0 (b)

(a) L

L

=

60°

+

H

60°

F1

F2

(c) FIGURE 12.10 (a) Retaining wall in Design Example 12.5, (b) wall showing horizontal and vertical seawater forces, and (c) the wall broken into triangular and rectangular segments.

(4) The acceleration of gravity may be taken as g = 9.81 m/s2 . (5) The density of water may be taken as 1000 kg/m3 . We designate w = 1000 kg/m3 ,

sw = 1.025w ,

and

wall = 1.700w

as the densities of water, seawater, and earth-filled wall, respectively. We then determine the horizontal and vertical forces due to the seawater on the slanted surface of the retaining wall. These forces are shown in Figure 12.10b. The horizontal force will be  Fx = 0

H

 y2  H H2 sw gy dy = 1.025w g  = 1.025w g 2 0 2

( N)

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Introduction to Thermal and Fluid Engineering

and with H = 7.5 m Fx = 1.025w g

(7.5 m) 2 = 28.83w 2

(N)

The vertical (downward) seawater force is due to the weight of the seawater above the slanted surface between A and B. With a volume of water of   H cos 60◦ H2 (7.5 m) 2 V= H= = = 14.06 m3 2 4 4 we have F y = sw gV = 1.025w g(14.06 m3 ) = 14.41w g

(N)

The next step is to determine the moments due to the seawater. With clockwise moments taken as positive, the moment about point O produced by Fx will be   H 7.5 m Mzx = Fx = (28.83w g) = 72.07w g (N-m) 3 3 A counterclockwise moment is produced by the weight of the water     2H ◦ Mzy = F y L + cos 60 = F y (L + 2.5 m) 3 or Mzy = (14.41w g) (L + 2.5 m)

(N-m)

The earth-filled retaining wall may be broken up into two component sections as indicated in Figure 12.10c. For F1 , we have the volume (for a wall depth of 1 m)   1 1 H2 1 (7.5 m) 2 ◦ V = ( H cos 60 ) H = = = 14.06m3 2 2 2 2 2 and F1 = 1.70w gV = 1.70w g(14.06m3 ) = 23.91w g The counterclockwise moment is   H cos 60◦ Mz1 = F1 L + = 23.91w g(L + 1.25 m) 3

(N)

(N-m)

Finally, for F2 and a wall depth of 1 m V = H L = (1 m)(7.5 m)(1 m) = 7.5 m3 F2 = 1.70w gV = 1.70w g(7.5L) = 12.75w g and the counterclockwise moment is     L L Mz2 = F2 = 12.75w g = 6.375w gL 2 2 2  We now take moments about point 0 ( M = 0) Mzx − Mzy − Mz1 − Mz2 = 0

(N)

(N-m)

Fluid Statics

377

and with the substitution of the calculated moments 72.07w g − (14.41w g)(L + 2.5 m) −(23.91w g)(L + 1.25 m) − 6.375w gL 2 = 0 Cancelation of the common w g terms and simplification gives L 2 + 2.26(L + 2.5 m) + 3.75(L + 1.25 m) − 11.30 = 0 or L 2 + 6.01L − 0.96 = 0 and the solution to this quadratic gives the result sought L = 0.155 m

12.6

(15.5 cm) ⇐

Hydrostatic Forces on Curved Surfaces

We next examine the forces exerted on a submerged plane surface in a more general sense than was considered in Section 12.4. The simplest approach to the evaluation of the force exerted on the top of the curved surface, AB, shown in profile in Figure 12.11a, is to consider a free-body diagram of the fluid in the column ABCD above the surface shown in Figure 12.11b. It is clear that the net horizontal force exerted on the surface, (F2 in Figure 12.12b), is the pressure force exerted by the fluid on a vertical projection of surface AB. Moreover, it is apparent that the vertical force on surface AB is merely the weight of a column of fluid extending from the surface AB to the free-liquid surface. The vector sum of these two components will be the total hydrostatic force exerted on the curved surface AB. Evaluation of these two components is straightforward. There may be some mathematical complexity if the surface has an odd configuration.

Example 12.5 Suppose that the curved surface, AB, in Figure 12.11 is a quarter circle of radius 50 cm while BD, the vertical surface, is 40-cm high. Both surfaces are 20-cm deep (into the plane of the paper). Determine the horizontal and vertical components of the hydrostatic force acting on surfaces AB and BD per unit width if the surfaces are in contact with water having a density of 1000 kg/m3 .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluid is incompressible. (3) The acceleration of gravity is assumed to be 9.81 m/s2 . (4) The density of the water is specified. With reference to Figure 12.11b, the horizontal component, F1 , on the surface BD is F1 = Pcg S = g y¯ S = (1000 kg/m3 )(9.81 m/s2 )(0.20 m)(0.20 m)(1 m) = 392.4 N Because the surface, BD, is vertical, the vertical component of the hydrostatic force is zero.

378

Introduction to Thermal and Fluid Engineering C

D

C

D

F1

F1 W1

B

B

F2

F2

W2

A

A (a)

(b)

FIGURE 12.11 (a) A submerged curved surface shown in profile and (b) a free-body diagram of the fluid in the column ABCD above the surface.

The horizontal component, F2 , on the surface, AB, is based on the projected surface area, Sproj F2 = g y¯ Sproj = (1000 kg/m3 )(9.81 m/s2 )(0.40 m + 0.25 m)(0.50 m)(1 m) = 3188.3 N ⇐ The vertical component, W1 + W2 , is the weight of the column of water standing above the surface, AB W1 + W2 = (1000 kg/m3 )(9.81 m/s2 )     × (0.5 m) 2 + (0.50 m)(0.40 m) (1 m) = 3888.2 N ⇐ 4

12.7

Buoyancy

The well-known principle of Archimedes, dating from the third century BC, applies to a body that is totally immersed in a fluid: An immersed body is buoyed up by a buoyant force that is equivalent to the weight of the fluid displaced by the body. If the body is less dense than the fluid in question, the buoyant force will be equal to the product of the fluid density and the volume of the displaced fluid. In this case, the body will float because the weight of the volume of fluid required to equal the weight of the body is less than the weight of the total volume of the body. The buoyant force is clearly a manifestation of the hydrostatic forces exerted by the fluid on the surfaces of an immersed object. This can be visualized quite readily by an example.

Example 12.6 A concrete block lies, half-buried, at the bottom of a lake that is 7 m deep. The density of the concrete is 2310 kg/m3 and the block is rectangular measuring 1 m by 1 m

Fluid Statics

379

F

F y

7m

7m

h WC WC (b)

(a)

FIGURE 12.12 The concrete block for Example 12.6 (a) originally half-buried and (b) just lifted free.

by 0.15 m. The 0.15-m side is the one that is half-buried. Determine (a) the force required to lift the block free from the bottom of the lake and (b) once pulled free, the force necessary to maintain its position (Figure 12.12).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is no flow of fluid. (3) The fluid is incompressible. (4) The acceleration of gravity is assumed to be 9.81 m/s2 . (5) The density of the water is 1000 kg/m3 . (a) To obtain the force required to lift the block free, we write a force balance in the vertical direction. The required freeing force, F, is the weight of a column of water, Ww , that is 6.925 m deep above the top surface of the block plus Wc , the weight of the block itself F = Ww + Wc Here, with w = 1000 kg/m3 and the top surface area equal to 1 m2 , we can write Ww = w gSy =

(1000 kg/m3 )(9.81 m/s2 )(1 m2 )(6.925 m) 1 kg-m/s2 -N

= 67.93 kN and Wc = c gSh =

(2310 kg/m3 )(9.81 m/s2 )(1 m2 )(0.15 m)

= 3.40 kN

1 kg-m/s2 -N

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Hence, F = Ww + Wc = 67.93 kN + 3.40 kN = 71.33 kN ⇐ (b) If we consider the block to be at some arbitrary distance, y, below the liquid surface, the hydrostatic forces on the top and bottom surfaces with S = 1 m2 will be Ftop = Ptop S = w gyS and Fbot = Pbot S = w g( y + h)S The net force, F , required to maintain this position will be F = Ftop − Fbot + Wc = w gyS − w g( y + h)S + c gSh = −w gh S + c gSh The first of these two terms is seen to be equal to the weight of water contained in the volume Sh and is the upward or buoyant force described by Archimedes. Thus, F =−

(1000 kg/m3 )(9.81 m/s2 )(1 m2 )(0.150 m) 1 kg-m/s2 -N

+ 3.40 kN

= −1.47 kN + 3.40 kN = 1.93 kN ⇐ We note that this result is independent of the depth, y, and is a striking example of the effect of buoyancy that is really an imbalance in hydrostatic pressure around an immersed body.

12.8

Stability

We are not only concerned with the force balance on a body that is floating or immersed in a fluid but also with its stability. We are able to evaluate the static stability of a body by perturbing or disturbing its position slightly and then seeing whether or not the forces on it tend to return it to its original position. If the body will return it is said to be stable. If, when perturbed, it continues to move toward another equilibrium position it is statically unstable. We are familiar with unstable situations such as a pencil standing on its point, a clown standing on his head, and a thin coin standing on edge. For a body totally immersed in a fluid, such as the balloon and its passenger shown in Figure 12.13, the situation is stable as long as the center of buoyancy (point B) is above the center of gravity (point G). If the balloon system is displaced from its stable configuration in Figure 12.13a by, say, a gust of wind, it will assume the position illustrated in Figure 12.13b. The moment produced when the weight and buoyant forces are not in line will tend to bring the balloon back to the position in Figure 12.13a, which is the stable configuration. When a body floats at the interface of two fluids, such as at the surface of a body of water, the circumstances are less obvious. Consider, for example, a rectangular solid block

Fluid Statics

381

W

W

B

B

G

G

W W (a)

(b)

FIGURE 12.13 The stability of a balloon system (a) the stable configuration and (b) the perturbed system.

floating in water, as indicated in Figure 12.14a, with its weight and its buoyant force in line. We note that the weight acts through point G, the center of gravity of the block, and the buoyant force acts through point B, the center of buoyancy, which is at the center of gravity of the displaced volume. Clearly, point G is above point B but the block may still be stable. Our analysis now involves the perturbed state shown in Figure 12.14b, where the two forces are not in line due to a displacement of the block. The weight of the block can still be considered a concentrated force through point G. If the buoyant force continued to act through point B, the block would continue to tumble and would represent an unstable condition. The tricky part now is to observe that, due to the tipping, volume Oab in Figure 12.14b, which was previously submerged, is now above the water and volume Ocd, which was formerly above the water, is now submerged. The result of both of these changes is that, compared to the original upright case, there is a greater buoyant force to the left of the centerline and a lesser one to the right. The result of this is that the new center of buoyancy, B  , is to the left of the previous position at B. If it is far enough to the left, then the condition is a stable one and the block will right itself. However, if the new center of buoyancy is less than some required amount, the condition will remain unstable.

W

C

B

O M G W

W

B'

G

(a)

D

A B

B

(b)

FIGURE 12.14 The stability of a floating block: (a) the stable configuration and (b) the perturbed system.

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To determine the stability of the new arrangement, we locate the new center of buoyancy, point B , and draw a vertical line through this point intersecting the symmetry axis of the block at point M. It is clear that if point M is above point G the block will be stable. Point M is designated as the metacenter and the distance MG is the metacentric height. The condition for stability is that point M be above point G or that the metacentric height be positive. Example 12.7, which now follows, should aid in illustrating this concept.

Example 12.7 A freighter is idealized as having a rectangular cross section as shown in Figure 12.15. It has a draft height of 2 m and a width of 2L. We wish to find the limiting ratio of L/ h for which this vessel will be stable. The center of gravity may be assumed to be at the waterline as shown.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is no flow of fluid. (3) The fluid is incompressible. The vessel is shown in the tipped configuration in Figure 12.15b. Because the cross section is symmetrical about the original vertical axis (Figure 12.15a) and rectangular, we see that the areas Ga b and Gcd are equal. If we take moments about point G, the limiting condition  will be for the case where M = 0. Proceeding, we may write this balance of moments for small displacement angles . We note that there are two new forces for the two new triangular volumes above and below the waterline. These two forces are located at a distance 2L/3 from point G which, as Figure 12.15 shows, is the distance to the centroid of each of the triangular volumes being considered. Taking moments about point G gives 

1 Lcd 2



    2 1 2 L + La b L − 2Lh(G B) = 0 3 2 3

The distances cd, a b, and G B may be approximated for small displacements as cd = a b = L L a G

d B

L

h

Δθ

G

c

b L B

L (a)

(b)

FIGURE 12.15 The cross section of the freighter of Example 12.7: (a) the stable configuration and (b) the perturbed system.

Fluid Statics

383

and GB =

h  2

After substitution of these into the moment equation, we obtain       1 2 2 1 2 2 h L L + L L − 2Lh = 0 2 3 2 3 2 This yields 2 2 L − h2 = 0 3 or L = h

 1/2 3 = 1.225 ⇐ 2

This is the lower limit of the ratio, L/ h for the ship to be stable. If the ship is made wider, that is, if L > 1.225h, the stability will increase accordingly.

12.9

Uniform Rectilinear Acceleration

Recall that in Section 12.1, we made the distinction between inertial and noninertial reference frames. We will now consider the special noninertial case of a coordinate system that is accelerating uniformly. Newton’s law for such a system equates the sum of all of the forces to the mass acceleration product  F = ma (12.17) A development that precisely follows the steps in Section 12.1 would lead, with the fluid at rest relative to a system with constant acceleration, to the expression −∇ P + g = a

(12.18)

An alternate form that is consistent with our fundamental fluid statics relation of Equation 12.2 is ∇ P = (g − a)

(12.19)

Equation 12.19 states that the maximum rate of change of pressure is in the g − a direction. If the magnitude of the acceleration, a = 0, Equation 12.19 reverts to Equation 12.2. Equation 12.19 indicates that the fluid surface in a tank car, initially at rest, as shown in Figure 12.16a will assume the configuration shown in Figure 12.16b when the tank and its contents are accelerated to the right with magnitude, a . As indicated by the vector triangle in Figure 12.16b, the fluid surface, which will be at constant pressure will be perpendicular to the direction of g − a. At point b, for example, an analysis of the case shown will yield dP y = −|g − a|y = −(g 2 + a 2 ) 1/2 y dz where |g − a| indicates the magnitude of g − a.

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y

g

g

V

V h

x (g – a)

b

a

g –a (a)

(b)

FIGURE 12.16 A uniformly accelerated fuel tank (a) at constant velocity and (b) under uniform acceleration.

Separation of the variables and integration between y = 0 and y = −h will give the result  Pb  −h 2 2 1/2 d P = −(g + a ) dy Patm

0

or Pb − Patm = (g 2 + a 2 ) 1/2 h

(12.20)

A common example of uniform acceleration is the case of a spinning container that contains a liquid at atmospheric pressure. Example 12.8, which now follows, should aid in illustrating this concept.

Example 12.8 Determine the shape of the free surface of the liquid in the rotating cylindrical container shown in Figure 12.17. The container, with inside radius, r , spins with a uniform angular velocity, .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluid is incompressible.

ω R

z

h

g r ω

(a) FIGURE 12.17 The spinning cylinder containing a liquid of Example 12.8.

(b)

Fluid Statics

385

(3) The angular velocity, , is specified. (4) The acceleration of gravity is assumed to be 9.81 m/s2 . We will employ a cylindrical coordinate system and, for a constant , we may presume that the free-liquid surface will be symmetric about the centerline. Thus, the pressure will be a function of the coordinates r and z and we can write ∂P

dP =

∂r

dr +

∂P ∂z

dz

As we have noted, the pressure is constant along the free surface so that d P = 0 and the differential equation for such a constant pressure case is dz ∂ P/∂r =− dr ∂ P/∂z The necessary values for these partial derivatives may be obtained from the fluid statics relationship of Equation 12.19, ∂P ∂r

∂P

r+

∂z

z = (g − a)

The gravitational acceleration is constant, that is g = −gz and the radial acceleration is a function of r and may be expressed as a = −r 2 r With these in hand, we may write Equation 12.18 as ∂P ∂r

r+

∂P ∂z

z = −gz + r 2 r

We now equate the coefficients of the unit vectors to obtain ∂P ∂r

= r 2

and ∂P ∂z

z = −g

These are the terms required to solve for the shape of the free-liquid surface, y(r ). We may make the appropriate substitutions and obtain dz r 2 = dr g and the normal separation of the variables and integration yields  y  2 r dz = r dr g 0 h y−h =

2r 2 2g

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Introduction to Thermal and Fluid Engineering

The equation of the free surface is therefore, y=h+

2 r 2 ⇐ 2g

which is a parabola with its vertex at the point h on the cylinder centerline.

12.10

Summary

In this chapter we have examined the behavior of static fluids, which are motionless relative to a coordinate system that is either motionless or accelerating uniformly. The basic equation known as the fundamental equation of fluid statics is ∇ P = (g − a)

(12.19)

where the acceleration of the coordinate system is either zero or a constant. The usual situation is where a = 0 in which case ∇ P = g

(12.2)

Thus, fluid statics deals with problems that are associated with fluids at rest. A common pressure measuring device, the manometer, is based on principles of fluid statics. This chapter also discussed static situation forces on submerged surfaces, buoyancy, and stability. The stability of floating and immersed objects has been examined. A totally immersed system is stable as long as the center of mass is below the center of buoyancy. On the other hand, a floating object can be stable, unstable, or neutrally stable depending on the metacentric height, defined as the distance between the center of mass and the center of buoyancy. The floating body, such as the hull of a vessel, will be stable as long as the metacenter is above the center of buoyancy (as long as the metacentric height is positive).

12.11

Problems

General Considerations 12.1: Using standard conditions to determine the density of air, determine the height of the atmosphere if it were incompressible. 12.2: The isothermal compressibility of a substance is given by  = ( ∂ P/∂) T = −V( ∂ P/∂V) T . Determine  for a perfect gas. 12.3: In water, the modulus, , defined in Problem 12.2, is nearly constant and has a value of 2.068 × 106 kPa. Determine the percentage volume change in water due to a pressure of 20,000 kPa. 12.4: On a certain day, the barometric pressure at sea level is 30 in of mercury and the temperature is 20◦ C. The pressure gage in an aircraft in flight indicates a pressure of 73 kPa and a temperature gage shows the air to be at 8◦ C. Estimate the altitude of the aircraft above sea level. 12.5: If the density of seawater is approximated by the equation  = o e ( P−Pa tm )/

Fluid Statics

387

where o is the density at atmospheric pressure and  is the compressibility, take o = 1025 kg/m3 and  = 2.068 × 106 kPa and determine the pressure and density at a point 9750 m below the surface. 12.6: The practical depth limit for a suited diver is about 18 m. If the specific gravity of seawater is 1.025, determine the gage pressure in the seawater at that depth. 12.7: The deepest known point in the ocean is 11,034 m in the Mariana Trench in the Pacific. Assuming seawater to have a constant density of 1050 kg/m3 , determine the pressure at this point in atmospheres. 12.8: Determine the depth change to cause a pressure increase of 1 atm for (a) water, (b) seawater (SG = 1.0250), and (c) mercury (SG = 13.6). Hydrostatic Forces 12.9: A tank is filled with glycerin (SG = 1.258) to a level of 1.625 m. If the air space above the glycerin is pressurized to 40 kPa, determine the pressure at the bottom of the tank. 12.10: A tank is filled with mercury (SG = 13.6) such that the pressure on the bottom of the tank is 112.04 kPa. Determine the height of the mercury. 12.11: An open tank contains 0.50 m of oil with a specific gravity of 0.825 on top of 2.65 m of water. The barometer reads 29.72-in Hg. Determine (a) the pressure at the oil-mercury interface and (b) the pressure at the bottom of the tank. 12.12: A column of 1.25 m of lubricating oil (SG = 0.885), 2.875 m of water, and 6.48 m of a heavy oil yields a pressure of 228.47 kPa at the bottom of the column. If the atmospheric pressure is 100.23 kPa, determine the specific gravity of the heavy oil. 12.13: Mercury (SG = 13.6) is poured in the column of fluid considered in Problem 12.4 so that the absolute pressure at base of the column becomes 473.59 kPa, determine the height of the mercury in the column. 12.14: In Figure P12.14, the fluid is water and the left-hand side is open to an atmospheric pressure of 100 kPa. Determine the pressure PA. PA

100 kPa

3.625 m Water FIGURE P12.14

12.15: In Figure P12.15, the left-hand side is open to the atmosphere at 1.013 bar. Determine the pressure in the air space (a) above column B and (b) above column C.

B

C

1m 3m

1m Water FIGURE P12.15

3m

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Introduction to Thermal and Fluid Engineering

12.16: In Figure P12.16, PA = 200 kPa. Determine the value of PB . A

150 cm

B

Oil SG = 0.82

Water

75 cm

100 cm

100 cm

Mercury FIGURE P12.16

12.17: In the closed tank of Figure P12.17, gage A reads 305 kPa. Determine (a) the value of h and (b) the reading on gage B. 140 kPa Water

h

82.5 cm 6.5 cm

Mercury B

A

6.5 cm

FIGURE P12.17

12.18: A hydraulic jack is sketched in Figure P12.18. Determine the force, F , required to hold the 225-kg mass in place. 9000 N 7.5 cm diameter 2.5 cm 40 cm

F

2.5 cm diameter Oil FIGURE P12.18

Barometers and Manometers 12.19: In Figure P12.19, determine the absolute pressure at point A if the liquid in the manometer has a specific gravity of 1.55 and the barometric pressure is 100 kPa.

SG = 1.55

2m

A FIGURE P12.19

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389

12.20: The tank shown in Figure P12.20 is closed and the air pressure above the oil is 128.23 kPa. The manometer is open to the atmosphere and the barometric pressure is 29.92 in of mercury. Determine the height, h, of the oil in the manometer.

Air 135 kPa

0.625 m

h Oil SG = 0.85

0.625 m 1.625 m 1.0 m

Water

FIGURE P12.20

12.21: The manometer shown in Figure P12.21 is open to the atmosphere. Determine the gage pressure at point A. Kerosene SG = 0.86 30.5 cm

A 18 cm

Gage Oil SG = 1.54 FIGURE P12.21

12.22: A pipe carrying water is tapped by a manometer that is open to the atmosphere as shown in Figure P12.22. Determine the gage pressure at point A.

A 4 cm Mercury

Water

16 cm

FIGURE P12.22

12.23: The tube shown in Figure P12.23 is open to the atmosphere at both ends. Determine the specific gravity of the oil.

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Introduction to Thermal and Fluid Engineering Patm

Patm

0.42 m 0.36 m

Water FIGURE P12.23

12.24: A differential manometer is connected at points A and B to a vertical pipe as shown in Figure P12.24. Determine the pressure difference between points A and B. Mercury

80 cm 40 cm

B

Water

2.0 m

A

FIGURE P12.24

12.25: In Figure P12.25, determine the mass of the piston if the pressure gage reads 124.85 kPa.

124.85 kPa

50 cm 87.5 cm

W Oil SG = 0.845

FIGURE P12.25

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391

12.26: In Figure P12.26, tank 1 contains water and tank 2 contains a more dense liquid. Both tanks are open to the atmosphere. For the configuration shown, what is the value for the specific gravity of the oil? 1 Water 1.80 m 2

Oil

1.20 m 37.5 cm

Mercury FIGURE P12.26

12.27: In Problem 12.26, determine the level of the water that will cause the pressure difference to be zero. 12.28: A mercury manometer, open to the atmosphere, is connected to a pipe containing oil as indicated in Figure P12.28. Determine the gage reading. Pressure Gage

1.50 m

4.50 m

Oil SG = 0.875

0.24 m Mercury FIGURE P12.28

12.29: In Figure P12.29, point B is at 200 Pa. Liquids 1 and 2 have densities of 857 kg/m3 and 1261 kg/m3 , respectively. Determine the pressure at point A. Liquid 1

Liquid 2 40 cm

1.96 m Mercury A

3.0 m

B FIGURE P12.29

PB = 200 kPa

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12.30: Determine the difference in pressure between points A and B in Figure P12.30.

B

0.16 m

30° Air

A Water

0.48 m 0.32 m

Mercury FIGURE P12.30

12.31: The specific gravity of the oil in Figure P12.31 is 0.80 and the manometer is open to the atmosphere. Determine the pressure at point A.

A Oil SG = 0.80

3.2 m

1.0 m Water

34 cm

Mercury FIGURE P12.31

12.32: In Figure P12.32, determine the height of the oil indicated by h if both manometers are open to the atmosphere.

h

30 cm

12 cm

Oil SG = 0.845 Water

30 cm FIGURE P12.32

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393

12.33: A differential manometer is used to measure the pressure change caused by a flow constriction in a piping system as shown in Figure P12.33. Determine (a) the pressure difference between points A and B (Pa) and (b) which section has the higher pressure. H2O B 7.5 cm A Hg

2.5 cm

FIGURE P12.33

12.34: Determine the pressure at point A in Figure P12.34.

Water at 65°C

Kerosene 17.5 cm

A 5 cm 12.5 cm

Mercury FIGURE P12.34

12.35: In the system shown in Figure P12.35, all fluids are at 20◦ C. Determine the pressure difference between points A and B. Kerosene

40 cm

Air A Benzene

20 cm

8 cm

15 cm

10 cm Mercury

B

Water FIGURE P12.35

12.36: In the system shown in Figure P12.36, fluid 1 is glycerin and fluid 2 is carbon tetrachloride and Pa = 101 kPa. Determine the absolute pressure at point A.

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Introduction to Thermal and Fluid Engineering Pa

ρ1 35 cm

A 10 cm

ρ2 FIGURE P12.36

12.37: The U-tube manometer shown in Figure P12.37 has a 1 cm-ID and contains mercury. Determine the free-surface height in each leg at equilibrium if 10 cm3 of water is poured into the right-hand leg. 1 cm

12 cm

8 cm FIGURE P12.37

12.38: The hydraulic jack shown in Figure P12.38 is filled with a hydraulic fluid having a density of 865 kg/m3 . Neglect the weight of the two pistons and determine what force on the handle is required to support the 100-N weight. 100 N 7.5 cm diameter 2.5 cm 40 cm

F

2.5 cm diameter Oil FIGURE P12.38

Forces on Plane Areas 12.39: Freshly poured concrete approximates a fluid with a specific gravity of 2.40. Figure P12.39 shows a 2.5 m×3 m×0.200 m slab that is poured between two wooden forms, which are connected by four corner bolts. Neglect end effects and compute the forces in all four bolts.

Fluid Statics

395 Concrete B 3.00 m

A D C

2.44 m 20 cm FIGURE P12.39

12.40: Glass viewing windows are to be installed in an aquarium. Each window is to be 0.6 m in diameter and centered 2 m below the surface level. Determine the magnitude and location of each force acting on the window. 12.41: Determine the minimum value of h for which the gate shown in Figure P12.41 will rotate counterclockwise if the gate cross section is triangular, 1.2 m at the base, and 1.2-m high. Neglect bearing friction.

H2O

41.40 kPag

h 1.20 m

Air Gate (a)

(b)

FIGURE P12.41

12.42: A dam spillway gate holds back water of depth, h, as shown in Figure P12.42. The gate weighs 6000 N and it is hinged at A. Determine the depth of water for which the gate will raise up and permit water to flow under it. 4.60 m CG h

A 60°

1.80 m

3.00 m FIGURE P12.42

12.43: A watertight bulkhead, 6.75-m high, forms a temporary dam for some construction work. As indicated in Figure P12.43, the top 3.75 m contains seawater with  = 1030 kg/m3 behind the bulkhead, and the bottom 3 m contains a mixture of mud and water that can be considered as a fluid having a density of 2100 kg/m3 . Determine the total horizontal load per unit width and the location of the center of pressure measured from the bottom.

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3.75 m Sea Water P = 1030 kg/m3

6.75 m

Mud & Water P = 2100 kg/m3

3.00 m

FIGURE P12.43

12.44: The circular gate, ABC, shown in Figure P12.44, has a 1-m radius and is hinged at B. Neglect atmospheric pressure and determine the force, P, just sufficient to keep the gate from opening when h = 12 m. Pa Water Pa h A 1m B 1m C

P

FIGURE P12.44

12.45: As indicated in Figure P12.45, gate AB is semicircular, hinged at B, and held by a horizontal force, P at A. Determine the force, P, required for equilibrium.

6m Water A 4m B

P Gate: Side View FIGURE P12.45

12.46: Gate AB in Figure P12.46 is hinged at point A located 3.2 m above the surface of the water on its left. The gate is 3.6-m wide and the height of the water is 10.8 m. A

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397

volume of seawater (SG = 1.023), to a depth of h m, is on the right-hand side of the gate. Determine h, in meters, to just permit the gate to open.

A

3.2 m

Hinge

Water 10.8 m

Sea Water h SG = 1.023

FIGURE P12.46

12.47: The gate, AB, in Figure P12.47 is hinged at B and has a weight of 9000 N. It rests against a smooth wall at A. Water is present on both sides; on the left to an unknown height, h, and on the right to a depth of 14 m. The gate is 6-m wide. Determine the height, h, which will just cause the gate to open.

Water

h 6m

A 10 m 8m Gate B Hinge FIGURE P12.47

12.48: The vertical plate in Figure P12.48 is submerged vertically in water. Determine the magnitude of the hydrostatic force on one side and the depth to the center of pressure.

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Introduction to Thermal and Fluid Engineering

2m

Water 4m Plate

S2 S1

4m

4m

4m FIGURE P12.48

12.49: Gate AB in Figure P12.49 is hinged at B and, when opened, rotates in the clockwise direction. The width of the gate is 2 m. Determine the counterclockwise torque required to hold the gate shut.

4m A

10 m

B Hinge

FIGURE P12.49

12.50: To control the water surface in a reservoir, a hinged leaf gate having the dimensions indicated in Figure P12.50 is provided. The gate is shown in its upper limiting position and presses on the seat at the end of the bottom leaf. If the weight of the gate is neglected, determine the magnitude of the force (per linear meter of crest) that is exerted at the seat in a direction normal to the gate. 5.5 m 4.25 m 1.20 m

Hinge 4.88 m

DAM

FIGURE P12.50

12.51: A tank that is 2.5-m wide contains 4 m of oil (SG = 0.780) and 2.4 m of water as indicated in Figure P12.51. Determine the magnitude and depth of the total force on side ABC.

Fluid Statics

399

4m

Oil SG = 0.780

2.4 m

Water FIGURE P12.51

12.52: Figure P12.52 shows an open triangular channel in which the two sides, AB and AC, are held together by cables, spaced 1 m apart, between B and C. Determine the cable tension. Cable

Water

1m

8m

6m FIGURE P12.52

12.53: The dam shown in Figure P12.53 is 100-m wide. Determine the magnitude and location of the force on the inclined surface.

Water at 27°C

128 m

96 m FIGURE P12.53

12.54: A vertical rectangular gate, having a width of 3 m, is located in a wall with water on the right-hand side as shown in Figure P12.54. Determine the force on the gate and its location.

3m

2m

Water

Gate

FIGURE P12.54

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Introduction to Thermal and Fluid Engineering

12.55: Repeat Problem 12.54 using an isosceles triangular gate with a base length of 2 m and with the vertex facing downward. Forces on Curved Areas 12.56: A dam structure has a rounded shape at its bottom as shown in Figure P12.56. The dam width is 50 m. Neglecting the effect of atmospheric pressure, determine the horizontal and vertical forces on the dam structure and determine their line of action. 8m Water 8m

FIGURE P12.56

12.57: Neglecting atmospheric pressure and assuming the width of the curved surface AB in Figure P12.57 to be 4 m, determine the total hydrostatic force exerted on the surface, AB. A 4m

Water

B FIGURE P12.57

12.58: A quarter circle gate, AB, of radius 0.70 m, is hinged at B as indicated in Figure P12.58. The concave surface of the gate is wetted by a fluid of density, 920 kg/m3 . Determine the force necessary to hold the gate stationary. B Fluid ρ = 920 kg/m3

0.70 m

A FIGURE P12.58

12.59: A 4-m-wide gate in the form of a quarter circle radius 3 m holds back water to a level of 1.8 m as shown in Figure P12.59. Determine the magnitude and direction of the resultant hydrostatic force. 3m 1.8 m

FIGURE P12.59

12.60: The flow from a water reservoir is controlled by a sluice gate 3-m wide. The gate is in the form of a sector of a circle with a radius of 1.2 m and an angle of 18◦ as shown

Fluid Statics

401

in Figure P12.60. Determine the magnitude and direction of the resultant force on the gate.

0.6 m

1.2 m 18°

FIGURE P12.60

12.61: As shown in Figure P12.61, a cylinder having a radius of 1.4 m holds back, on its left half, a fluid with a specific gravity of 0.84. The cylinder has a length of 2.5 m and is made of a material with a density of 145 kg/m3 . Determine the forces, F1 and F2 , that must be applied to hold the cylinder stationary.

Fluid SG = 0.84

1.4 m F1 ρ = 145 kg/m3

F2 FIGURE P12.61

12.62: As indicated in Figure P12.62, in the flat side of a tank holding water, there is a hemispherical surface of diameter, d, whose centerline is located h below the water surface. Derive expressions for the horizontal force, Fh , and the upward force, Fu , exerted on the hemispherical surface by the surrounding water.

h d Water FIGURE P12.62

12.63: In Figure P12.63, compute the magnitude of the horizontal and vertical components of hydrostatic force acting on the quarter circle of radius 0.75 m and width 1.5 m. The density of the oil is 850 kg/m3 .

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Introduction to Thermal and Fluid Engineering

Oil 2m

0.75 m ρ = 850 kg/m3 FIGURE P12.63

Buoyancy 12.64: The open end of a cylindrical tank that is 1 m in diameter and 1.5-m high is submerged in water as shown in Figure P12.64. The local barometric pressure is 101.3 kPa. The tank weighs 750 N and the thickness of the tank walls may be neglected. To what depth, h, will the tank submerge? 1.00 m

1.50 m h

FIGURE P12.64

12.65: In Problem 12.64, determine (a) the depth at which the net force on the tank is zero and (b) the additional force required to bring the top of the tank flush with the water surface. 12.66: A 12 m diameter balloon (Figure P12.66) is filled with helium (molecular weight 4.00) and is pressurized to 105 kPa. The surrounding air is at 100 kPa and 20◦ C. Determine the tension in the mooring line.

d = 12 m

Air 100 kPa 20°C

FIGURE P12.66

12.67: The hydrometer shown in Figure P12.67 floats at a level that is a measure of the specific gravity of the liquid. The stem is of constant diameter, d, and a weight in the bottom bulb stabilizes the body. The total hydrometer weight is 0.375 N and the

Fluid Statics

403

stem diameter is 0.08 m. Determine the height at which the device floats when the liquid has a specific gravity of 1.20.

d = 8 mm SG = 1.0 h

Fluid SG = 1.20 W

FIGURE P12.67

12.68: A barge weighs 360,000 N empty and is 8-m wide × 16-m long × 2.25-m high. Determine the depth below the waterline when loaded with 1.335 × 106 N of gravel floating in seawater (SG = 1.025). 12.69: A cube of side, L, and density, , floats in water with density, w , as shown in Figure P12.69. Show that the depth of submergence, h, is given by h=

L w

L

L h Water, pw FIGURE P12.69

12.70: The float in a toilet tank is a sphere of radius R and is made of a material with density, . An upward buoyant force, F , is required to shut the ball cock valve. The density of water is designated w . Develop an expression for the fraction of the float submerged, x, in terms of R, , F, g, and w . 12.71: An elliptical cylinder has a major axis of 10 cm, a minor axis of 4 cm, and a length of 50 cm. Its density of  = 545 kg/m3 . The cylinder is kept submerged in a water container with the help of a string attached to the bottom of the container as shown in Figure P12.71. Assuming the density of water to be 1000 kg/m3 , determine the tension, T, in the string.

404

Introduction to Thermal and Fluid Engineering

4 cm

10 cm

Water

FIGURE P12.71

12.72: A rectangular block of dimensions 12 cm by 8 cm is 80-cm long. The block is held submerged in water with a string and a horizontal pivoted rod 4 cm long, as shown in Figure P12.72. The mass in the block is not evenly distributed and consequently its center of mass is 2 cm from its geometric center. Assuming that the density of water is 1000 kg/m3 , the density of the block is 500 kg/m3 , g = 9.81 m/s2 and that the mass of the horizontal rod is negligible, determine the tension, T, in the string.

Pivot

Water

12 cm

8 cm

CM 2 cm

4 cm

FIGURE P12.72

12.73: As shown in Figure P12.73, a rectangular block with sides, a and b, length, L, and density, 3 , floats at the interface between fluids of density, 1 and 2 . Show that the ratio, b 1 /b 2 is given by b1 3 − 2 = b2 1 − 3 a Liquid 1 ρ1

b1

Liquid 2 ρ2

b2

FIGURE P12.73

12.74: Figure P12.74 shows a cylinder of 25-cm diameter and 75-cm height, resting on the floor of a container filled with a liquid of density,  = 800 kg/m3 . The density of the cylinder is 250 m3 . Determine the magnitude of the force, F.

Fluid Statics

405

25 cm

25 cm

75 cm

Cylinder ρ = 250 kg/m3

Liquid ρ = 800 kg/m3

F

FIGURE P12.74

12.75: Rework Problem 12.74 with the cylinder replaced by a right circular cone with a base diameter of 25 cm and a height of 75 cm. 12.76: A spherical balloon that is 8 m in diameter is filled with helium ( = 0.15 kg/m3 ). A cage having a mass of 15 kg capable of carrying a load of 160 kg is attached to the bottom of the balloon. The balloon is released in the air ( = 1.2 kg/m3 ) with its full load. Determine the net upward force exerted on the balloon. 12.77: The truncated cone shown in Figure P12.77 is completely immersed in a fluid of density,  = 760 kg/m3 . Determine the density of the truncated cone if the cone is balanced by a pulley-string arrangement that carries a mass of 5 kg.

9 cm 5 kg 20 cm

Liquid p = 760 kg/m3 12 cm FIGURE P12.77

Stability 12.78: A cubical piece of wood with specific gravity, 0.90, and edge length, L, floats in water with the right-hand edge of the block flush with the water as shown in Figure P12.78. What moment, M, is required to hold the cube in the position shown? L L M SG = 0.90

FIGURE P12.78

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Introduction to Thermal and Fluid Engineering

12.79: A cube measuring 50 cm on a side and made of a material with a density of 400 kg/m3 floats partially submerged in saltwater ( = 1050 kg/m3 ). Determine the metacentric height. 12.80: A cylinder of radius, R, and length, L, floats in water as shown in Figure P12.80. Show that when the cylinder is in neutral equilibrium, the metacentric height is zero.

R

θ θ

R

Water FIGURE P12.80

12.81: Figure P12.81 shows a cylinder of radius, R, and height, H, floating in water. The specific gravity of the cylinder is S. Show that the metacentric height, MG, is given by   1 R2 MG = − H(1 − S) 2 2H S and that for stable equilibrium R > [2S(1 − S)]1/2 H R S

G

H

Water

FIGURE P12.81

12.82: A rectangular barge is 16-m long, 6-m wide, and 2-m high. The barge has a mass of 80,640 kg when empty. Given that the density of seawater is 1050 kg/m3 , determine H, from the waterline to the bottom of the barge when it is floating.

13 Control Volume Analysis—Mass and Energy Conservation

Chapter Objectives •

To apply the law of mass conservation to fluid mechanics.

•

To apply the first law of thermodynamics to fluid mechanics.

To provide applications of fluid mechanics to the control volume expression for the first law of thermodynamics. • To introduce the Bernoulli equation. •

13.1

Introduction

Fluid motion is ubiquitous in the world around us. There are countless situations in which the understanding and harnessing of fluid motion provide the means for improving our lives. The heating and cooling of our homes and workplaces, the operation of engines for power and transportation and the transport of working fluids through pipelines and channels are a few examples. Our task in this and the next chapter is to develop a set of concepts that will allow a systematic examination of fluids in motion and their related effects. This is a logical progression beyond the fluid statics considerations examined in the previous chapter.

13.2

Fundamental Laws

All analyses of fluids in motion involve applications of three fundamental physical laws that are •

The law of mass conservation (continuity)

•

The first law of thermodynamics (energy)

•

Newtons’s second law of motion (momentum)

Continuity, energy, and momentum are terms that are often used when referring to a specific law. For example, the first law of thermodynamics is often referred to as the energy equation. We will use these designations interchangeably throughout this text; such usage is commonplace in the fluid mechanics community. 407

408

Introduction to Thermal and Fluid Engineering

We will deal exclusively with incompressible flows in this text. The analysis of compressible flows, which is beyond our scope, requires the application of the second law of thermodynamics in addition to the three laws just cited. There are also a number of other useful relationships that are often referred to as “laws.” Such relationships are more properly designated as constitutive relationships because they generally define a fluid property. One of these was introduced in Chapter 11: Newton’s viscosity relationship, which some treatises call Newton’s “law” of viscosity. In this text, we will reserve the term “law” for one of the three fundamental laws.

13.3

Conservation of Mass

The mass balance for a control volume was developed in Chapter 5 and expressed in operational form as Equation 5.1, which is restated here for reference   mcv = m ˙i − m ˙e (13.1) dt e i In words, Equation 13.1 states that the rate of mass accumulation in a control volume is equal to the total rate of mass inflow entering, (subscript i) minus the total rate of mass outflow exiting (subscript e).

13.4

Mass Conservation Applications

Equation 13.1 is completely general. It will be instructive to now consider some specific situations in which this expression may be applied. Two important cases will be considered initially: incompressible flow and steady flow. Incompressible flow, as explained earlier, is flow in which the changes in density are so negligibly small that the density may be treated as constant. Keep in mind that we may often treat the flow of a compressible fluid, such as air, as incompressible. For the case of incompressible flow, Equation 13.1 takes the form 

 dV  = i Ai Vˆ i − e Ae Vˆ e dt e i

or  dV  ˆ = Ai V i − Ae Vˆ e dt e i

(13.2)

where we have treated the density as constant. In steady flow, there is no variation of conditions with time. Operationally, this means that time derivatives vanish. Thus, Equation 13.1 reduces to   m ˙i − m ˙e =0 (13.3) i

e

and for the situation where the flow is both steady and incompressible   Ai Vˆ i − Ae Vˆ e = 0 i

e

(13.4)

Control Volume Analysis—Mass and Energy Conservation

409 Control Volume

ρ2, V2, A2

ρ3, V3, A3

Section 3

Section 2

Section 1

ρ1, V1, A1 FIGURE 13.1 Flow through a T section in Example 13.1.

We must also keep in mind that Equations 13.3 and 13.4 do not mean that there is no flow but merely that the rates of inflow and outflow are the same: that is, there is zero accumulation. We now consider three illustrative examples.

Example 13.1 Consider the T section shown in Figure 13.1. Flow enters at section 1 and is then split into two streams, which leave at sections 2 and 3. Express the mass balance for this T section under steady flow conditions.

Solution Assumptions and Specifications (1) The control volume is enclosed by dashed lines in Figure 13.1. (2) Steady-state conditions exist. (3) The flow is incompressible. (4) Uniform properties exist at each cross section where fluid enters or leaves the control volume. Because this is a condition of steady flow, the applicable mass conservation expression for a control volume is Equation 13.3   m ˙i − m ˙e =0 (13.3) i

e

We will apply this equation to the control volume shown in Figure 13.1 (dashed lines). Note that mass crosses the control surface at only three places, at sections 1, 2, and 3. Evaluation of the flow terms provides  m ˙ i = 1 Vˆ 1 A1 i



m ˙ e = 2 Vˆ 2 A2 + 3 Vˆ 3 A3

e

and, in accordance with Equation 13.3, we have 1 Vˆ 1 A1 = 2 Vˆ 2 A2 + 3 Vˆ 3 A3

(13.5)

410

Introduction to Thermal and Fluid Engineering Section 1

Section 2

R

r

r V2 = V max 1 – R A2 ρ

V1 = V avg A1 ρ

2

FIGURE 13.2 Steady incompressible flow in a circular pipe in Example 13.2.

This result states that the mass flow in at section 1 in the control volume is equal to the sum of mass flows out at sections 2 and 3. If, in addition, the flow were incompressible, we would have 1 = 2 = 3 and our result would have the form Vˆ 1 A1 = Vˆ 2 A2 + Vˆ 3 A3

(13.6)

which is in terms of the volume flow rates in and out. In Example 13.1, we have used average values for the densities at the three sections. This is certainly allowable if these values are known. Example 13.2 will relate local and average values for a specific case.

Example 13.2 Water enters a straight section of pipe from a reservoir as shown in Figure 13.2. At some distance downstream from the entrance, the velocity profile is known to have a parabolic form expressed as   r 2  ˆ ˆ Vav = Vmax 1 − (13.7) R ˆ max is the value of velocity at the centerline of the pipe (at r = 0). From the no-slip where V condition, the velocity is zero at the pipe wall where r = R. For the case of incompressible ˆ av or the average velocity in terms of V ˆ max . steady flow, we are to evaluate the value for V

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The water is incompressible. (3) Uniform properties exist where the fluid enters the pipe. For steady incompressible flow, the applicable form of the law of conservation of mass is Equation 13.4. Our control volume is enclosed by the dashed line in Figure 13.2. For this ˙ = AVˆ case, we can evaluate Equation 13.4 in terms of the volumetric flow rate, V   ˙i − ˙e = 0 V V i

e

Control Volume Analysis—Mass and Energy Conservation

411

At section 1, with Vˆ 1 = Vˆ avg , we have  ˙i = V ˙ avg = A1 V ˆ av = R2 V ˆ av V i

and at section 2 where we have the parabolic velocity distribution 

˙e = V ˙2 = V

 0

e

R

  r 2  ˆ max 1 − V 2r dr R

or  r3 r − 2 dr R 0  2 R r r4 ˆ max = 2V − 2 4R2 0

˙ 2 = 2V ˆ max V

ˆ max = V



R

R2 2

When we put these expressions for the volumetric flow rate at sections 1 and 2 together, we achieve the result ˆ max V Vˆ 1 = Vˆ avg = 2

(13.8)

Flow measurements are made by a variety of devices. The basic ideas of these devices will be presented in a later chapter. One of these, the pitot tube, measures the velocity at a point. An approximation for the velocity profile across a pipe or duct can then be made and the total flow evaluated as illustrated in Example 13.3.

Example 13.3 A pitot tube that traverses a circular duct having an inside diameter of 50 cm yields the following velocity values: Distance from Center, cm

Velocity, m/s

Distance from Center, cm

Velocity, m/s

0 8.3 11.3 13.9 16.1 18.0

2.29 2.16 2.06 1.96 1.87 1.77

19.7 21.3 22.7 24.1 25.0

1.67 1.55 1.37 1.16 0.73

From these data find (a) the flow rate in m3 /s and (b) the average velocity.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at the cross section of interest.

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Introduction to Thermal and Fluid Engineering Section 1

Section 2

R

++ ++ + +

r

V2

V1 = Vavg A1 ρ

A2 = ΣΔAi i ρ

FIGURE 13.3 Steady incompressible flow in a circular duct in Example 13.2.

The information provided is in discrete form rather than in continuous form as in an equation. Nevertheless, the governing relationship remains the conservation of mass, Equation 13.1. For the case of steady, incompressible flow, Equation 13.1 reduces to   m ˙i − m ˙e =0 (13.3) e

i

This situation is indicated in Figure 13.3 with the velocity at section 2 represented as discrete points rather than a smooth curve as shown in Figure 13.2. The solution to our form of the mass conservation expression can be written as   ˙i − ˙e = 0 V V e

i

which reduces further to



Vˆ i Ai − Vˆ avg A = 0

i

or Vˆ avg A =



Vˆ i Ai

(13.9)

i

and finally Vˆ avg =

 i

Ai Vˆ i A

(13.10)

In Equation 13.9, the product Vˆ i Ai is a volumetric flow rate through a portion of the cross section, designated as Ai , where Vˆ i is the average velocity over the section, Ai . Referring now to the data provided in the problem statement, we may consider each of the velocity values to be Vˆ i , applying over area Ai as indicated in Figure 13.4. For example, the velocity, Vˆ i = 1.87 m/s, is associated with a segment of the total cross section lying between r1 =

13.9 cm + 16.1 cm 16.1 cm + 18.0 cm = 15.00 cm and r2 = = 17.05 cm 2 2

Control Volume Analysis—Mass and Energy Conservation

413

ΔAi r2

r1

FIGURE 13.4 Cross section of duct in Example 13.3.

The corresponding value for Ai is therefore, Ai = [(17.05 cm) 2 − (15.00 cm) 2 ] = 206.4 cm2

or

0.0206 m2

In a similar fashion, each Vˆ i can be related to Ai . A summary of this procedure is shown in the following table. Vˆ i , m/s

ΔAi , m2

Vˆ i ΔAi , m3 /s

2.29 2.16 2.06 1.96 1.87 1.77 1.67 1.55 1.37 1.16 0.73

0.0050 0.0242 0.0206 0.0208 0.0206 0.0203 0.0204 0.0200 0.0200 0.0173 0.0070

0.0115 0.0523 0.0424 0.0408 0.0385 0.0359 0.0341 0.0310 0.0274 0.0201 0.0051



Ai = 0.1962 m2

ˆ V i Ai = 0.3391 m3 /s

pjwstk|402064|1435600722

The total flow rate is seen to be ˙ = 0.3391, m3 /s ⇐ V and the average velocity is given by 

Vˆ i Ai

i

Vˆ avg =  i

Ai

=

0.3391 m3 /s = 1.73 m/s ⇐ 0.0192 m2

414

13.5

Introduction to Thermal and Fluid Engineering

The First Law of Thermodynamics for a Control Volume

The governing control volume relationship for the first law of thermodynamics was developed in Section 5.4.4 as Equation 5.11. This relationship may be rewritten here as  ˆ2  V d E cv ˙ cv + ˙ cv − W =Q m ˙ i h i + i + gzi dt 2 i   Vˆ 2 − m ˙ e h e + e + gze 2 e

(13.11)

Equation 13.11 will be the starting point for all subsequent analyses in this chapter. The reader is reminded that d E cv /dt is the rate of increase of the total energy within the control volume, ˙ cv is the net rate of heat transfer into the control volume from the surroundings, • Q ˙ cv is the net rate of work done by the control volume on its surroundings, • W •

•

m ˙ i is the mass flow rate into the control volume by means of inlet stream, i,

h i = ui + Pi /i is the enthalpy of the inlet stream, i, ˆ 2 /2 is the kinetic energy of the inlet stream, i, and • V i •

•

gzi is the potential energy of the inlet stream, i.

Each of the terms written with the subscript, i, has a counterpart with the same meaning for the exiting streams bearing the subscript, e. To review the specific energy terms, observe that in Equation 13.11, the energy per unit mass of a quantity of fluid flowing into or out of a control volume may be written in terms of its component parts as e+

P Vˆ 2 P = gz + +u+  2 

(13.12)

P Vˆ 2 = gz + +h  2

(13.13)

or e+

where gz is the specific potential energy, Vˆ 2 /2 is the specific kinetic energy, h = u + P/ is the specific enthalpy, and u is the specific internal energy. The reader is cautioned at this point to observe that P/ and the enthalpy, h, represent valid energy terms only at a flow boundary and that the accumulation term does not include these components. Equation 13.11 now assumes its place as one of the principle control volume equations that will form a basis for fluid flow analysis. All quantitative descriptions of fluid flow to be considered in this text are based on this expression, along with Equation 13.1 and the yet-to-be-developed expressions for linear momentum and moment of momentum. The use of Equation 13.11 and the interpretation of the various terms that it contains will be clarified in the next section where a number of practical examples will be considered in detail.

Control Volume Analysis—Mass and Energy Conservation

13.6

415

Applications of the Control Volume Expression for the First Law

A series of examples is presented in this section to provide the reader with some insights concerning the energy equation.

Example 13.4 For the nozzle shown in Figure 13.5, we wish to know the magnitude of the change in internal energy of the flow. This can be thought of as the energy loss to friction that is rendered unavailable to do mechanical work. The upstream pressure in the pipe is 530 kPa (absolute).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at the cross section at the inlet and outlet of the nozzle. (4) Atmospheric pressure may be taken as 101 kPa. Applying Equation 13.11 to the control volume indicated in Figure 13.5, we first note that d E cv /dt = 0 so that ˙ cv = W ˙ cv = E e − E i Q ˙ cv = W ˙ cv = 0, and with both Q 

 m ˙i

i

Vˆ 2 h i + i + gzi 2 

=m ˙1

−



 m ˙e

e

Vˆ 2 h e + e + gze 2



 Vˆ 21 Vˆ 22 P1 P2 u1 + + + gz1 − m ˙ 2 u2 + + + gz2 = 0  2  2 Section 1

Section 2

r Flow

Atmospheric Pressure

x

Control Volume P1 = 530 kPa (Abs) d1 = 5 cm FIGURE 13.5 Water flowing through a horizontal nozzle in Example 13.4.

d2 = 2.5 cm

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Introduction to Thermal and Fluid Engineering

Because mass conservation dictates that m ˙1 =m ˙ 2 , our expression becomes u1 − u2 +

Vˆ 2 − Vˆ 22 P1 − P2 + 1 + g(z1 − z2 ) = 0  2

(13.14)

The quantity sought is u2 − u1 , which can be expressed as u2 − u1 =

Vˆ 2 − Vˆ 22 P1 − P2 + 1 + g(z1 − z2 )  2

The terms on the right-hand side of this expression are evaluated as follows: P1 − P2 530 kPa − 101 kPa = = 429 m2 /s2  1000 kg/m3 ˙2  1 Vˆ 21 − Vˆ 22 V 1 = − 2 2 2 A21 A2

 2  2  (53 m3 /h) 2 4 4 = − 2(3600 s/h) 2 (0.050 m) 2 (0.025 m) 2 = −422 m2 /s2 and g(z1 − z2 ) = 0 The resulting value of u2 − u1 thus becomes u2 − u1 =

Vˆ 2 − Vˆ 22 P1 − P2 + 1 + g(z1 − z2 )  2

= 429 m2 /s2 − 422 m2 /s2 = 7 m2 /s2 ⇐ This result indicates that 7 m2 /s2 of energy is lost due to frictional effects and other dissipative processes in the nozzle. It is reassuring that the answer is positive. A negative result would have suggested that dissipative processes are increasing the useful energy potential of the leaving stream that is clearly not physically possible. The units of this result may be confusing at first glance. The terms in Equation 13.14 are all appropriate forms of energy per unit mass although they may not seem so. The units for gz, the potential energy change and for Vˆ 2 /2, the kinetic energy change, are in m2 /s2 . The internal energy terms and the flow energy change are in N-m/kg. The conversion is straightforward in this example  2 s -N 2 2 7 m /s = 7 N-m/kg 1 kg-m This conversion is relatively simple in SI units where the numerical value of gc is unity, The reader is hereby alerted to the situation where unit conversions are likely to be a major source of effort and potential error in using the energy equation. One should be careful in the process of obtaining a numerical result in a specific set of units.

Control Volume Analysis—Mass and Energy Conservation

417

Section 2 Section 1

Fan Air Flow

A

0.35 m ?

3.2 cm

B

C H2O

Section 3

FIGURE 13.6 Induced-draft fan and duct assembly in Example 13.5.

Example 13.5 provides another illustration of the use of the control volume expression of the first law of thermodynamics.

Example 13.5 Consider the fan and duct system shown in Figure 13.6. The duct has a diameter of 0.35 m and its entrance is smoothly rounded to minimize losses. Air is drawn through the duct by the induced-draft fan and the indicated differential manometer shows a vacuum pressure of 3.2 cm of water. The density of air may be taken as 1.22 kg/m3 . Determine (a) the volumetric flow rate of the air (m3 /s) and (b) the power required to drive the fan (kW).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The liquid in the manometer is incompressible. (3) Uniform properties exist at each cross section of interest. (4) No change in internal energy will be assumed. (a) We choose a control volume that will allow the direct determination of the volumetric flow rate. Because the power required to drive the fan is not known, we should not include the fan in our control volume. The dashed lines indicate an appropriate choice for the control volume. We know the conditions at both sections 1 and 2 and, because nothing has been given concerning u, the frictional losses, we will neglect frictional effects. The steady flow form of the energy equation applies and we have g(z2 − z1 ) +

Vˆ 22 − Vˆ 21 P2 − P1 + u2 − u1 + =0 2 

At section 1 P1 = Patm

and

Vˆ 1 = 0

and at section 2 P2 = Patm − 3.2 cm H2 O

and

˙ V Vˆ 2 = A2

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Introduction to Thermal and Fluid Engineering

Then Equation 13.14 u1 − u2 +

Vˆ 2 − Vˆ 22 P1 − P2 + 1 + g(z1 − z2 ) = 0  2

(13.14)

gives us with, u = z and Vˆ 1 = 0  ˙ 2 Vˆ 2 1 V P1 − P2 = = 2 2 A2  Then with 1 cm of water = 98.1 N/m2 and with A2 =

 (0.35 m) 2 = 0.0962 m2 4

we write  1/2 ˙ = A2 2( P1 − P2 ) V     1/2 2[Patm − ( Patm − 3.2 cm water)] 98.1 N/m2 1 kg-m = A2 1 cm water N-s2 1.22 kg/m3  = A2

2(3.2 cm water) 1.22 kg/m3



98.1 N/m2 cm water

 1

1 kg-m

1/2

N-s2

= 22.69A2 m/s Thus, ˙ = (22.69 m/s) A2 = (22.69 m/s)(0.0962 m2 ) = 2.18 m3 /s ⇐ V (b) To establish the power to drive the fan, we need to define a control volume that includes it. The choice made, as indicated in Figure 13.6, is between sections 1 and 3. For the case of steady flow, we employ Equation 13.11 with d E cv /dt = 0 ˙ cv − W ˙ cv + Q

 i

 m ˙i

Vˆ 2 h i + i + gzi 2

−

 e

 m ˙e

Vˆ 2 h e + e + gze 2



The summation terms may be rearranged to the form  i

 m ˙i

Vˆ 2 h i + i + gzi 2

−

 e

 m ˙e

Vˆ 2 h e + e + gze 2



  V 2 − V32 P1 − P3 =m ˙ (u1 − u3 ) + + 1 + g(z1 − z3 )  2

=0

Control Volume Analysis—Mass and Energy Conservation

419

where m ˙ =m ˙1 =m ˙ 3 . Now, with ˙ cv = 0 Q Vˆ 1 = 0 u1 − u3 = 0 P1 = P3 = Patm z1 − z3 = 0 ˙ cv = −W ˙ fan , the expression to be solved is and W   2  ˙ 1 V ˙ fan = −m W ˙ 2 A3 and with A3 = A2 , this yields  2 3 3 2.18 m3 /s ˙ fan = − (1.22 kg/m )(2.18 m /s) W 2 (/4)(0.35 m) 2 = −683 W = −0.683kW ⇐ The negative sign indicates that the power is transmitted from the surroundings to the control volume. Because the fan is a part of the control volume, the power to drive the fan is the unknown quantity sought, and thus the negative sign makes sense.

13.7

The Bernoulli Equation

The energy equation, under certain conditions, reduces to a form that is referred to as the Bernoulli equation. This equation is sufficiently important and useful that we devote an entire section to its discussion. The following conditions apply: ˙ =0 Adiabatic flow (no heat transferred), Q ˙ =0 • No work done, W •

•

Steady flow

•

Incompressible flow

•

Negligible viscous effects (inviscid flow)

Application of the first four of the foregoing conditions yields the steady-flow form of the energy equation, Equation 13.14, which was derived in a Example 13.5. An added condition for this case is that of inviscid flow, which gives u = 0. The resulting form of the control volume expression for energy is the Bernoulli equation gz +

Vˆ 2 P + =C 2 

(a constant)

(13.15)

The Bernoulli equation establishes that, for the five conditions specified, the total energy in a fluid is manifested as the sum of potential, kinetic, and flow energies. Any change in one of these forms requires appropriate changes in the others.

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1

V = 20 m/s

d = 0.10 m

2 H2O

ρair = 1.21 kg/m3

h

FIGURE 13.7 A pitot-static tube in an air duct.

As expressed by Equation 13.15, the units of each term are in m2 /s2 . If each term is divided by g, the result, given in meters is z+

Vˆ 2 P + =C 2g g

(a constant)

(13.16)

In this form, each energy term is designated as a “head” due to elevation (potential energy), velocity (kinetic energy), and pressure (flow energy). Two examples involving the Bernoulli equation will now be considered. A pitot static or pitot tube is employed in Example 13.6 to measure velocity. In Figure 13.7, the two pressure lines are situated differently relative to the flow. At point 1, because of the orientation of the tube opening with respect to the flow, the fluid is brought to rest. The pressure at point 1 is designated the stagnation pressure or impact pressure. At point 2, the orientation of the tube opening is such that it does not obstruct the flow, and the pressure sensed at point 2 is known as the static pressure.

Example 13.6 Air flows in a duct with pressure probes is attached to a manometer as shown in Figure 13.7. If the density of air is 1.21 kg/m3 and that of water as 1000 kg/m3 determine the elevation difference, h, between the two legs of the manometer.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section. We select the centerline of the duct as the reference elevation and note in Figure 13.7 that the difference in elevation between points 1 and 2 is half of the duct diameter, or 0.05 m. An application of the Bernoulli equation at both points yields the relation z1 +

Vˆ 21 Vˆ 2 P1 P2 + = z2 + 2 + 2g g 2g g

Noting that Vˆ 1 = 0 (at the stagnation point) and that z2 − z1 = −0.05 m, we obtain Vˆ 2 P1 − P2 = z2 − z1 + 2 g 2g = −0.05 m + = 20.3 m

(20 m/s) 2 2(9.81 m/s2 )

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421

3

0

Nozzle Diameter d5 = 5.08 cm 6.7 m

Water 4

5

7.6 m ˆ V 5 2 Pipe Diameter 10.16 cm

1

FIGURE 13.8 A water tank and discharge system in Example 13.7.

This is the pressure head that is required to bring the velocity to zero but is not the manometer displacement. The manometer will register the same pressure difference in units of meters of water. Hence,  P 1.21 kg/m3 of air = 20.3 m of air  1000 kg/m3 of water = 0.0246 m of water

or

2.46 cm of water ⇐

The use of pitot tubes to evaluate velocity is common and we will see them again. Usually, the velocity is evaluated by reading the displacement of fluid in the manometer. By positioning the impact pressure probe at a number of positions across the flow passage, the series of local velocities measured will provide a velocity profile of the flow. This was the case considered in Example 13.3.

Example 13.7 An open water tank and piping system is shown in Figure 13.8. The water discharges downstream through a 5.08-cm-diameter nozzle. Assuming that the liquid level in the tank remains constant and neglecting frictional losses, determine (a) the rate of water discharge through the nozzle and (b) the pressure and velocity at points 1, 2, 3, 4, and 5.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. For the conditions specified in the problem statement, the Bernoulli equation applies everywhere and can be employed to relate conditions between the various points. (a) At the outset, consider the surface of the fluid in the tank at point 0, where Vˆ 0 = 0 and the exit from the nozzle at point 5. Both locations are at atmospheric pressure.

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Equation 13.16 gives z0 +

Vˆ 2 Patm Patm = z5 + 5 + g 2g g

from which the exit velocity from the nozzle can be calculated as Vˆ 5 = [2g(z0 − z5 )]1/2 = [2(9.81 m/s2 )(6.7 m)]1/2 = (131.45 m2 /s2 ) 1/2 = 11.47 m/s ⇐ (b) Because Vˆ 5 = 11.47 m/s is the velocity through the 5.08-cm nozzle, the velocity through the 10.16-cm-diameter pipe will be  Vˆ = 11.47 m/s

5.08 cm 10.16 m/s

2 = 2.87 m/s

This is the velocity at all other points of interest Vˆ 1 = Vˆ 2 = Vˆ 3 = Vˆ 4 = 2.87 m/s Between points 0 and 1, we have z0 +

Vˆ 20 Vˆ 2 Patm P1 + = z1 + 1 + 2g  2g 

which, with Vˆ 0 = 0, gives Vˆ 2 P1 − Patm = z0 − z1 − 1  2g = 7.6 m −

(2.87 m/s) 2 2(9.81 m/s)

= 7.18 m of water This is the gage pressure at point 1. We note, by inspection, that conditions at points 2 and 4 are identical because the velocities and elevations are the same at these points. The pressure at point 2 may be determined by applying the Bernoulli equation between points 0 and 2. z0 +

Vˆ 20 Vˆ 2 Patm P2 + = z2 + 2 + 2g  2g 

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423

which with Vˆ 0 = 0 gives Vˆ 2 P2 − Patm = z0 − z2 − 2  2g = 6.70 m −

(2.87 m/s) 2 2(9.81 m/s)

= 6.28 m of water This is the gage pressure at points 2 and 4. The final unknown is the pressure at point 3. We again apply the Bernoulli equation, this time between points 0 and 3, and obtain z0 +

Vˆ 20 Vˆ 2 Patm P3 + = z3 + 3 + 2g  2g 

which, with Vˆ 0 = 0, gives Vˆ 2 P3 − Patm = z0 − z3 − 3  2g = 0.00 m −

(2.87 m/s) 2 2(9.81 m/s)

= 0.00 m − 0.42 m = −0.42 m of water This is the gage pressure at point 3. At point 3, the pressure is 0.42 m of water vacuum, that is, the pressure is below atmospheric pressure. It is clear that, as the part of the piping system containing point 3 is elevated, the pressure at the point will continue to decrease. It is of interest to think about the maximum height that might be reasonable. Is it possible for the absolute pressure at point 3 to be negative? An application of the Bernoulli equation to the design of a drinking fountain now follows.

13.8

Design Example 6

A public drinking fountain is to be designed. The fountain employs a nozzle to discharge drinking water vertically upward to a level that is 4 cm above the nozzle discharge. The fountain configuration is shown in Figure 13.9. In the absence of viscous effects, determine (a) the pressure at the gage indicated to produce this condition and (b) the flow rate of the water achieved.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The fluid is incompressible.

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4 cm

2 2 cm 5 cm

3 3 cm

P3 g

FIGURE 13.9 Artist’s conception of a bubble drinking fountain.

(3) Uniform properties exist at each cross section. (4) The acceleration of gravity may be taken as 9.81 m/s2 . (5) The density of water may be taken as 1000 kg/m3 . Because the flow is considered incompressible, the Bernoulli equation may be applied between locations of interest. We will break the configuration into two control volumes as indicated in Figure 13.10. (a) In Figure 13.10, control volume A extends from the fountain discharge upward to the level where the vertical water jet velocity reaches zero. The Bernoulli equation between levels 1 and 2 is Vˆ 2 Vˆ 2 P1 P2 + gy1 + 1 = + gy2 + 2 1 2 2 2 Noting that Vˆ 1 = 0 and that P1 = P2 = Patm , this expression reduces to

pjwstk|402064|1435600702

g( y1 − y2 ) =

Vˆ 22 2

or Vˆ 22 = 2g( y1 − y2 ) = 2(9.81 m/s2 )(0.04 m) = 0.785 m2 /s2 Vˆ 2 = 0.886 m/s For control volume B, extending between levels 2 and 3, we may write Vˆ 2 − Vˆ 23 P2 − P3 + g( y2 − y3 ) + 2 =0  2

Control Volume Analysis—Mass and Energy Conservation

1

425

A

2 B

3

P3 g

FIGURE 13.10 Control volumes used in the design of a bubble drinking fountain.

The quantity sought, P3 , may now be expressed as Vˆ 2 − Vˆ 23 P3 − P2 = g( y2 − y3 ) + 2  2 We note that P3 − P2 = P3 − Patm = P3g and that, by continuity,  Vˆ 3 = Vˆ 2

A2 A3



 = Vˆ 2

d2 d3

2

Thus,   4  Vˆ 22 P3g d2 = g( y2 − y3 ) + 1−  2 d3    (0.886 m) 2 0.02 m 4 = (9.81 m/s )(0.05 m) + 1− 2 0.03 m 2

= 0.491 m2 /s2 + 0.315 m2 /s2 = 0.806 m2 /s2

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Introduction to Thermal and Fluid Engineering

giving P3g = (0.806 m2 /s2 ) = (1000 kg/m3 )(0.806 m2 /s2 ) = 806 N/m2 = 806 Pa ⇐ (b) The water flow rate is ˙ = A2 Vˆ 2 =  (0.02 m) 2 (0.886 m/s) = 2.78 × 10−4 m3 /s ⇐ V 4

13.9

Summary

The law of mass conservation and the first law of thermodynamics, in forms that are applicable to a control volume, have been considered in this chapter. The operational equations for these laws are the conservation of mass equation   mcv = m ˙i − m ˙e (13.1) dt e i and the first law of thermodynamics  ˆ2  V d E cv ˙ cv − W ˙ cv + =Q m ˙ i h i + i + gzi dt 2 i   Vˆ 2 − m ˙ e h e + e + gze 2 e

(13.11)

We considered several examples that involved applications of these expressions to some situations of practical importance. Under certain conditions, the energy equation reduces to the Bernoulli equation, which is broadly applicable to a number of physical situations gz +

Vˆ 2 P + =C 2 

(a constant)

(13.15)

The conditions for which the Bernoulli equation applies are as follows: ˙ =0 Adiabatic flow (no heat transferred), Q ˙ =0 • No work done, W •

•

Steady flow

•

Incompressible flow

•

Negligible viscous effects (inviscid flow)

13.10

Problems

Conservation of Mass 13.1: Oil with a specific gravity of 0.85 flows through a 7.62-cm-diameter pipe at a velocity of 3 m/s. Determine (a) the volumetric flow rate and (b) the mass flow rate.

Control Volume Analysis—Mass and Energy Conservation

427

13.2: A tapering pipe has an inside diameter of 30 cm at section 1 and 45 cm at section 2. Water flows through the pipe at a velocity of 5 m/s at section 2. Determine (a) the velocity at section 1, (b) the volumetric flow rate at section 1, (c) the volumetric flow rate at section 2, and (d) the mass flow rate. 13.3: Air flows through a 40-cm × 80-cm rectangular duct. The air is at 27◦ C and 125 kPa and it flows at a rate of 1.25 kg/s. Determine (a) the average velocity and (b) the volumetric flow rate. 13.4: A chamber has one inlet and two outlets. The inlet (port 1) has a diameter of 10 cm and the volumetric flow rate is 0.0525 m3 /s. One outlet (port 2) has a diameter of 7.5 cm and an average velocity of 10 m/s. At the other outlet (port 3) the diameter is 2.5 cm. Determine (a) the volumetric flow rate at port 3 and (b) the average velocity at port 3. 13.5: Water flows through a converging nozzle at a mass flow rate of 80 kg/s. The nozzle inlet (section 1) has a diameter of 24 cm, and the outlet (section 2) has a diameter of 8 cm. Determine the average velocity at (a) the inlet and (b) the outlet. 13.6: A 12.5-cm plunger is pushed at a rate of 8 cm/s into a tank filled with oil having a specific gravity of 0.775. The tank outlet has a diameter of 1.5 cm. Determine the mass flow of fluid at the outlet. 13.7: A tank has two inlets and one outlet. Water enters at port 1, which has a diameter of 3.75 cm, at a velocity of 15.5 m/s; at port 2 water enters with a volumetric flow rate of 0.0192 m3 /s. The water leaves port 3, which has a diameter of 5 cm. The level of water in the tanks remains constant. Determine the exit velocity. 13.8: Oil with a specific gravity of 0.862 flows through a 75-cm-diameter pipeline at a flow rate of 100 L/s. Determine (a) the velocity and (b) the mass flow rate. 13.9: Air enters a 20-cm-diameter pipe at a temperature of 22◦ C at a velocity of 70 m/s and a pressure 105 kPa. The pipe tapers to a diameter of 8 cm, and the air leaves at 667◦ C at a pressure of 1.4 MPa. Determine the exit velocity. 13.10: For the pipe arrangement shown in Figure P13.10, the volumetric flow rate in pipe 2 is 40% of the volumetric flow rate in pipe 1. Assume that the fluid is incompressible and determine the mean velocities and the volumetric flow rates in pipes 2 and 3. d2 = 16 cm

V2

d1 = 35 cm 2 V1 = 12 m/s

1 3 d3 = 22 cm

V 3

FIGURE P13.10

13.11: As indicated in Figure P13.11, water flows in a 0.6-m-diameter pipe with a parabolic profile given by  r2 ˆ V = 16 1 − (0 ≤ r ≤ 0.3) 0.09 ˆ is the velocity in m/s. The water where r is the radial coordinate in meters and V is discharged through a sudden contraction into a 0.20-m-diameter pipe. Determine

428

Introduction to Thermal and Fluid Engineering the average velocity in this pipe.

2 V = 16(1 – r ) 0.09

0.20 m 0.60 m

FIGURE P13.11

13.12: Oil with a specific gravity of 0.82 flows through a 10-cm-diameter pipe at 0.15 m3 /s. Assuming the density of water to be 1000 kg/m3 , determine (a) the average velocity in m/s and (b) the mass flow rate in kg/s. ˆ i = 5 m/s, Ai = 20 cm2 and 13.13: Air flows through a conical diffuser. At the inlet V 3 ˆ e , is 1 m/s, and e = 0.98 kg/m3 . i = 1.02 kg/m . The desired velocity at the exit, V Determine (a) the exit area needed to provide this exit density and (b) the mass flow rate through the diffuser. 13.14: A fluid of constant density flows in the passage formed by two parallel plates. Letting L be the separation distance between the plates and y the distance from the lower plate, derive an expression for the average velocity if the velocity distribution is ˆ = C1 y, (b) V ˆ = C2 y1/2 , and (c) V ˆ = C3 (L y − y2 ). given by (a) V 13.15: Assuming ideal gas behavior and that air is compressible, consider Figure P13.15, which shows a convergent-divergent duct. The flow is one dimensional and the diameter, pressure, and temperature at stations 1, 2, and 3 are d1 = 40 cm

P1 = 200 kPa

T1 = 800 K

d2 = 8 cm

P2 = 100 kPa

T2 = 450 K

d3 = 25 cm

P3 = 60 kPa

T3 = 320 K

ˆ1, V ˆ 2 , and V ˆ3. The airflow in the duct is 0.4 kg/s. Determine the average velocities, V

2 3

1 P1, T1, d1

P2, T2, d2

P3, T3, d3

FIGURE P13.15

13.16: As indicated in Figure P13.16, water enters one end of a perforated pipe (20 cm in diameter) with a velocity of 6 m/s. The discharge through the pipe wall is approximated by a linear profile and the flow is steady. Determine the discharge velocity ˆ V.

Control Volume Analysis—Mass and Energy Conservation

429 V

V = 6 m/s

V

0.5 m 6 V FIGURE P13.16

13.17: A water jet pump possesses coaxial streams as shown in Figure P13.17. The jet emerging from the inner tube of 2-cm diameter has a velocity of 25 m/s. The jet emerging from the coaxial tube with a 15-cm diameter has a velocity of 4 m/s. The two streams mix thoroughly and flow through the exit. Determine the average exit ˆe . velocity V Coaxial (not to scale) stream

15 cm

Ve

Jet

2 cm

FIGURE P13.17

13.18: Air, which may be taken as an ideal gas with a constant density of 1.05 kg/m3 , flows through a 5-cm-diameter pipe with a velocity profile of   r  ˆ = 12 1 − sin V (0 < r < 2.5 cm) 5 ˆ is the velocity in m/s and r is the radial coordinate in centimeters. Determine where V (a) the average velocity and (b) the mass flow rate. ˆ 1 = 5 m/s 13.19: The water tank shown in Figure P13.19 is being filled through section 1 at V 3 ˙ and through section 3 at V3 = 0.01 m /s; the water level, h, is constant. Determine ˆ2. V . V3 = 0.01 m3/s

3

1 V1 = 5 m/s

2

h

d1 = 5 cm

V2 d2 = 7 cm Water d FIGURE P13.19

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Introduction to Thermal and Fluid Engineering

13.20: In Figure P13.20, oil with a density of 850 kg/m3 flows at a velocity of 2 m/s in a pipe with diameter of 38 cm. A sudden contraction reduces the pipe diameter to 25 cm. The pipe then splits into branches of diameters 15 and 10 cm. The velocity in the 15-cm-diameter pipe is 3.2 m/s. Determine (a) the mass flow rates and (b) the average velocity in the 10-cm branch.

d3 V1

38 cm

=1

5c

m

V3

d2 = 25 cm

d4 = 10 cm

V4

FIGURE P13.20

13.21: Water flows steadily in the control volume shown in Figure P13.21 in three circular sections. Section 1 has a diameter of 1.60 cm and the flow is 0.028 m3 /s. Section 2 has a diameter of 0.60 cm at an average velocity of 1.4 m/s. Determine the average velocity and volumetric flow rate at section 3 where, the diameter is 8 mm. d2 = 6 mm d1 = 1.6 cm . V1 = 0.28 m3/s

2 1

3

d3 = 8 mm

FIGURE P13.21

13.22: As indicated in Figure P13.22, a fluid fills the space between two very long plates of length, 2L, which are separated by a distance b. The upper plate moves downward ˆ Determine the mass flow rate and maximum velocity for (a) uniform with at a rate V. exit velocity and (b) parabolic exit velocity. V

b x 2L FIGURE P13.22

13.23: In some wind tunnels, the test section wall is porous or perforated; fluid is sucked out to provide a thin viscous boundary layer. The wall in Figure P13.23 is 4-m long and

Control Volume Analysis—Mass and Energy Conservation

431

contains 800 holes of 6-mm diameter per square meter of area. The suction velocity ˆ s = 12 m/s, and the test section entrance velocity is V ˆ 1 = 45 m/s. out of each hole is V ◦ Assuming incompressible flow of air at 20 C and a pressure of 1 atm, determine (a) ˆ 0 , (b) the total wall suction volume flow, (c) V ˆ 2 , and (d) V ˆf. V

Test Section ds = 0.8 m

df = 2.5 m

do = 3 m

Uniform Suction

Vf

V2

V 1

V0

L=4m FIGURE P13.23

13.24: The velocities in a circular duct of 1-m diameter were measured as follows: Distance from Center, cm

Velocity, m/s

0.0 5.0 10.0 15.0 20.0 25.0

2.52 2.30 2.26 2.01 1.92 1.80

Distance from Center, cm

Velocity, m/s

30.0 35.0 40.0 45.0 50.0

1.50 1.31 1.02 0.84 0.70

Determine (a) the average velocity and (b) the volumetric flow rate in m3 /s. 13.25: Water is flowing through the circular conduit shown in Figure P13.25 with a velocity profile given by  r2 Vˆ = Vˆ max 1 − 2 R where Vˆ max = 3 m/s and R = 1.25 m. Determine the average water velocity in the 80-cm-diameter pipe.

r 2.5 m d = 80 cm FIGURE P13.25

13.26: Water enters the conical diffusing passage shown in Figure P13.26 with an average velocity of 4 m/s. The entrance cross-sectional area is 0.0925 m2 . Determine the exit area needed to reduce the velocity to 0.40 m/s.

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Introduction to Thermal and Fluid Engineering

Section (1) Section (2) V2 = 0.40 m/s

V1 = 4 m/s A1 = 0.0925 m2

FIGURE P13.26

13.27: Water enters the 10-cm-square channel shown in Figure P13.27 at a velocity of 3.2 m/s. The channel converges to a 5-cm square at the outlet section, which is cut at a 30◦ angle to the vertical. Determine the outlet water velocity and the total rate of flow if the mean velocity of the discharging water remains horizontal. 5 cm 10 cm 5 cm

30° 10 c

m FIGURE P13.27

13.28: Water flows steadily through a nozzle (Figure P13.28) at 60 kg/s. The diameters are 25 cm at the left (d1 ) and 8 cm at the right (d2 ). Determine the average velocities at sections 1 and 2 in m/s. 1 2

FIGURE P13.28

13.29: The hypodermic syringe shown in Figure P13.29 contains serum (SG = 1.02). Assume that there is no leakage past the plunger, which is pushed in steadily at 1.5 cm/s. Determine the exit velocity, Vˆ 2 in m/s. d1 = 2 cm d2 = 0.075 cm V2

FIGURE P13.29

13.30: Given a garden hose with a flow rate of of 1 L/s, how long would it take to fill to a depth of 2 m a cylindrical swimming pool having a diameter of 8 m?

Control Volume Analysis—Mass and Energy Conservation

433

The First Law of Thermodynamics 13.31: A jet of water is projected vertically with a velocity of 22.5 m/s. Neglect air resistance and determine how high the jet will rise. 13.32: A water jet is inclined upward at an angle of 36.87◦ from the horizontal. Neglecting air resistance, determine the velocity required to allow the jet to clear a 3.5-m wall at a distance of 20 m. 13.33: Water flows in an inclined pipe from point 1 where the diameter is 30 cm, the elevation is 12 m, and the pressure head is 8 m, to point 2 where the diameter is 60 m and the elevation is 8 m. The volumetric flow rate is 0.375 m3 /s. Assume that there are no losses and determine the pressure head at point 2. 13.34: A water jet is inclined upward at an angle of 36.87◦ from the horizontal. Neglecting air resistance, determine the velocity required to allow the jet to clear a 3.5-m wall at a distance of 20 m. 13.35: Water flows in an inclined pipe from point 1 where the diameter is 30 cm, the elevation is 12 m, and the pressure head is 8 m, to point 2 where the diameter is 60 cm and the elevation is 8 m. The volumetric flow rate is 0.375 m3 /s. Assume that there are no losses and determine the pressure head at point 2. 13.36: Seawater ( = 1025 kg/m3 ) flows through a pump at a flow rate of 0.14 m3 /s. The pump inlet is 0.25 m in diameter. At the inlet, the pressure is −15 cm of water. The pump outlet, which is 0.152 m in diameter, is located 1.8 m above the inlet, and the outlet pressure is 1.75 kPa. The inlet and exit temperatures are equal. Determine the power that the pump adds to the fluid. 13.37: A liquid is heated in a 15-m-long vertical tube with a constant diameter of 5 cm. The flow is upward and at the entrance the average velocity is 1 m/s, the pressure is 340 kPa, and the density is 1000 kg/m3 . The increase in internal energy is 200 kJ/kg. Determine the amount of heat added to the fluid. 13.38: For the convergent nozzle shown in Figure P13.38, assume that the work done is zero and that the nozzle is adiabatic and determine the change in the internal energy of the water.

V2

V1

1

2 FIGURE P13.38

13.39: As indicated in Figure P13.39, a fan draws air from the atmosphere through a 0.30-mdiameter round duct that has a smoothly rounded entrance. A differential manometer connected to an opening in the wall of the duct shows a vacuum pressure of 2.5 cm of water. Given the density of air (1.22 kg/m3 ), determine (a) the volumetric rate of airflow and (b) the work done by the fan.

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Introduction to Thermal and Fluid Engineering

Fan 30 cm 2.5 cm FIGURE P13.39

13.40: A liquid of density 800 kg/m3 flows through the diffuser shown in Figure P13.40. The mass flow rate of the liquid through the diffuser is 10 kg/s and the change of internal energy between the inlet and outlet is 45 m2 /s2 . Assuming steady flow with no external heat or work interactions, determine the outlet pressure P0 .

.

m = 10 hg/s ρ = 800 kg/m3

Pi = 150 kPa di = 5 cm

L

do = 20 cm

O FIGURE P13.40

13.41: Water flows through the reducer shown in Figure P13.41 at the rate of 0.05 m3 /s. The diameters of the reducer are 12 cm and 7 cm. The pressure drop across the reducer is 95 kPa and the difference in elevation is 0.4 m. Determine the change in internal energy of the water as it flows through the reducer.

z2 – z1 = 40 cm

3 /s m . = 0.05 r te V Wa

2 d2 = 7 cm 1 d1 = 12 cm FIGURE P13.41

13.42: Water flows through a valve in a horizontal constant area pipe such that the change in pressure is 5 kPa. Determine the change in internal energy of the water. 13.43: Air at 20◦ C flows into a 0.285 m3 reservoir at a velocity of 20 m/s. The reservoir temperature and pressure are 20◦ C and 96 kPa, respectively. Assume that the incoming air is at reservoir pressure and flows through a 20-cm-diameter pipe. Determine the rate of temperature increase in the reservoir. 13.44: Water flows through a 5-cm-diameter horizontal pipe at 2.25×10−3 m3 /s. Heat transfer can be neglected and frictional forces cause a pressure drop of 480 Pa. Determine the rate at which heat is added to the water.

Control Volume Analysis—Mass and Energy Conservation

435

The Bernoulli Equation 13.45: The pressurized tank shown in Figure P13.45 has a circular cross section of 1.8-m diameter. Oil is drained through a nozzle 5 cm in diameter in the side of the tank. The specific gravities of the oil and mercury are 0.85 and 13.6, respectively. Assuming that the air pressure is kept constant, determine how long it will take to lower the surface of the oil in the tank by 1.50 m. P = 10 cm Hg Oil 1.50 m

5 cm

1.80 m FIGURE P13.45

13.46: Water in an open cylindrical tank, 3 m in diameter, discharges to the atmosphere through a nozzle, 5 cm in diameter. Neglecting friction and any unsteadiness in the flow, determine the time required for the water in the tank to drop from a level of 8.52 m above the nozzle to the 1.20-m level. 13.47: Water flows in an inclined pipe from point 1 where the diameter is 30 cm, the elevation is 12 m, and the pressure head is 8 m. At point 2, the diameter is 60 m and the elevation is 8 m. The volumetric flow rate is 0.375 m3 /s. Assume that there are no losses and determine the pressure head at point 2. 13.48: A pipe carries water from point 1 where the diameter is 8 cm and the elevation above the datum is 4 m, to point 2 where the diameter is 4 cm and the elevation above the datum is 10 m. The volumetric flow rate is 0.0375 m3 /s, and the pressure at point 1 is 988 kPa. Assuming no frictional losses, determine P2 . 13.49: An inclined pipe, carrying oil with a specific gravity of 0.885, changes in diameter from 15 cm at section 1 to 46 cm at section 2. Section 1 is 4 m lower than section 2 and the volumetric flow rate is 0.142 m3 /s. The pressures at sections 1 and 2 are 90 and 60 kPa, respectively, and losses are considered to be negligible. Determine the direction of flow.

pjwstk|402064|1435600714

13.50: Water flows from an opening in the bottom of a reservoir through a 5-cm-diameter pipe to a discharge point below the reservoir surface as shown in Figure P13.50. The volumetric flow rate is 0.01875 m3 /s, and no head is lost in the system. Determine the difference in height, h, between the reservoir surface and the discharge.

Water h

5 cm FIGURE P13.50

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Introduction to Thermal and Fluid Engineering

13.51: A nozzle with a diameter of 5 cm is attached to a horizontal pipe with a diameter of 10 cm. Assuming that head losses are negligible and the pressure upstream of the nozzle is 480 kPa, determine the water jet velocity. 13.52: Consider Figure P13.52 and show that when the siphon is running continuously that (a) the lowest pressure depends on y1 + y2 and is located at point 3 and (b) that the ˆ 2 depends upon both y1 and g. exit velocity, V 3

y2 1 y1 2 FIGURE P13.52

13.53: Water flows from station 1 to station 2 in a pipe. Station 2 is 2.25-m lower than ˆ 1 = 2.4 m/s. P1 = 325 kPa, d1 = 10 cm, d2 = 5 cm, and no station 1. Assuming V losses, determine P2 . 13.54: A large tank possesses a small, well-rounded opening, 25 m below the surface of the water in the tank. Determine the velocity of the water passing through the hole. 13.55: Oil with a specific gravity of 0.875 flows from a tank through a 15-cm-diameter pipe. Point 1 at an elevation of 36 m is under an unknown air pressure. Point 2 at an elevation of 4.5 m is at the discharge. If head losses are negligible, determine the air pressure P1 necessary to discharge 0.625 m3 /s of oil. 13.56: A pump delivers 0.0375 m3 /s of oil, having a specific gravity of 0.875, through a 10cm suction pipe. The pressure at point 1, which is 1.25 m below the datum that runs through the centerline of the pump, is a vacuum of 200 mm of mercury. Determine the total energy head at point 1. 13.57: A horizontal air duct has an area reduction from 0.0375 m2 at station 1 to 0.0100 m2 at station 2. For a mass flow of air at m ˙ = 0.075 kg/s, with a density  of 0.0995 kg/m3 , determine the pressure change between the stations when frictional losses are neglected. 13.58: Oil with a specific gravity of 0.825 is flowing through a pipe of 3-cm inner diameter at a pressure of 100 kPa. The total energy, relative to a reference level 24 m below the centerline of the pipe is 225 m. Determine the flow rate of the oil. 13.59: Water in an open cylindrical tank, 3 m in diameter, discharges through a nozzle that is 5 cm in diameter. Neglecting losses, determine the time required for the water in the tank to drop from a level of 6 m above the nozzle inlet to a level of 1.5 m. 13.60: A venturi meter is a carefully designed constriction whose pressure difference is a measure of the flow rate in a pipe. A venturi like the one shown in Figure P13.60, is connected to a horizontal pipe of 20-cm diameter. The water that flows through one pipe may have a maximum flow rate of 8.07 m3 /s. The pressure head at the inlet for this flow is 18 m above atmospheric, and the pressure head at the throat must not be lower than 7 m below atmospheric. Between the inlet and the throat, there is

Control Volume Analysis—Mass and Energy Conservation

437

an estimated frictional loss of 10% of the difference in pressure head between these points. Determine the minimum allowable diameter of the throat.

20 cm

FIGURE P13.60

13.61: A horizontal venturi meter has a diameter of 8 cm at its inlet and 4 cm at its throat. The pressure difference between the inlet and the throat is 25 kPa. Determine the volumetric rate of water flow through the venturi when no losses need be accounted for. 13.62: In Figure P13.62, the flowing fluid is air at a density  of 1.04 kg/m3 and the manometer fluid is oil with a specific gravity of 0.827. Assuming no losses, compute the flow rate in m3 /s. d1 = 10 cm

d2 = 6 cm

8 cm

FIGURE P13.62

13.63: Using Bernoulli’s equation for steady incompressible flow with no losses, show that the volumetric flow rate through a venturi meter is related to the manometer reading, h by  A2 2gh(m − ) ˙ V=  4  d2 1− d1 where m is the density of the manometer fluid. 13.64: In Figure P13.64, the manometer fluid is mercury. Neglecting losses, determine the flow rate in the tube in m3 /s if the flowing fluid is (a) water with  = 1000 kg/m3 and (b) air with  = 1.02 kg/m3 .

8 cm

2.5 cm

FIGURE P13.64

13.65: Oil with a specific gravity of 0.900 flows downward through a vertical contraction as shown in Figure P13.65. The mercury manometer reading h is 12 cm. Determine

438

Introduction to Thermal and Fluid Engineering the volumetric flow rate for frictionless flow and state whether the actual flow rate is more or less than the value for frictionless flow. 300 mm

60 cm

h

100 mm FIGURE P13.65

13.66: Air of density 1.21 kg/m3 is flowing as shown in both parts of Figure P13.66, and ˆ = 15 m/s. Determine the readings on the manometers. V V

Air V 15 m/s

Oil H2O

S.G. = 0.86

(a)

(b) FIGURE P13.66

13.67: In Figure P13.67, the flowing fluid is air with a density  of 1.2 kg/m3 , and the manometer fluid is oil with a specific gravity, SG = 0.827. Assuming no losses, compute the flow rate in m3 /s. D1 = 10 cm

D2 = 6 cm

8 cm

FIGURE P13.67

13.68: Water flows through the vertical pipe in Figure P13.68. Is the flow in the pipe up or down?

Control Volume Analysis—Mass and Energy Conservation

439

H

h Mercury FIGURE P13.68

13.69: A liquid flows in the horizontal pipeline shown in Figure P13.69. The velocity is 0.65 m/s with a friction loss of 0.1375 m of flowing fluid. For a pressure head at B of 0.60 m, determine the pressure head at A.

PA

30 cm

0.60 m

A

B

15 cm

FIGURE P13.69

13.70: Water flows steadily upward in the vertical pipe shown in Figure P13.70. It is then deflected to flow outward with uniform radial velocity. Neglecting friction and assuming that the pressure at A is 70 kPa, determine the volumetric flow rate of the water through the pipe. 30 cm

1.25 cm

20 cm 1.55 m

A FIGURE P13.70

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14 Newton’s Second Law of Motion

Chapter Objectives •

To introduce the control volume relationship for linear momentum.

•

To provide applications of the momentum theorem.

•

To consider the control volume relationship for the moment of momentum.

•

To use applications taken from a number of physical situations to show applications of the moment of momentum relationship.

14.1

Introduction

The third fundamental law to be considered is Newton’s second law. As in Chapter 13, we will develop control volume expressions, which, in this case, will be related to both linear and angular motion. The basic expressions will then be applied to a number of physical situations.

14.2

Linear Momentum

Newton’s second law of motion may be stated as The time rate of change of momentum of a system is equal to the net force on the system and takes place in the direction of the net force. This statement is notable in two ways. First, it is cast in a form that includes both magnitude and direction and is, therefore, a vector expression. Second, it refers to a system rather than a control volume. As we know by now, a system is a fixed collection of mass, whereas a control volume is a fixed region in space that encloses a different mass of fluid (or system) at different times. The transformation of the second law statement from a system to a control volume point of view is dealt with in numerous texts. We will presume the correctness of the following word equation: ⎫ ⎫ ⎧ ⎧ ⎨ Net force ⎬ ⎨ Rate of momentum ⎬ out of the control acting on the = ⎭ ⎭ ⎩ ⎩ volume by mass flow control volume (14.1) ⎫ ⎫ ⎧ ⎧ ⎨ Rate of momentum ⎬ ⎨ Rate of accumulation ⎬ into the control + of momentum within − ⎭ ⎭ ⎩ ⎩ the control volume volume by mass flow 441

442

Introduction to Thermal and Fluid Engineering Control Volume

. me

. mi

FIGURE 14.1 A general control volume and flow field.

The control volume shown in Figure 14.1 has the same general features that are shown in Figure 13.1. The total force acting on this control volume, due to both surface and body  forces, will be treated as a single vector term F. As with our earlier control volume considerations, the rate of mass flow through any single “passageway” across the control surface is written as m ˙ i or m ˙ e with the subscripts indicating whether the mass flow is entering (subscript i) or exiting (subscript e). The momentum fluxes associated with these mass flows are expressed as Vˆ i m ˙ i and Vˆ e m ˙ e. ˆ ˆ The velocity vectors V i and V e are the velocities of the corresponding mass flows relative to the defined coordinate system. This coordinate system may be defined in whatever manner is most convenient in a given application and it may either be stationary or moving. This choice will become clearer when examples are considered in the next section. The accumulation term in Equation 14.1 may be written as ⎧ ⎫ ⎨ Rate of accumulation ⎬ d of momentum within = Vˆ cv mcv ⎩ ⎭ dt the control volume where the velocity vector, V cv , is the velocity at which the control volume is moving relative to the defined coordinate system. The resulting control volume expression for linear momentum can now be written as

F=



Vˆ e m ˙e −

e

i

d Vˆ i m ˙ i + Vˆ cv mcv dt

(14.2)

The momentum flux terms are now shown as the sums of the contributions of all entering and exiting streams. Equation 14.2 is often referred to as the momentum theorem. It is the third of our fundamental control volume relationships. The similarity between Equation 14.2 and its counterparts, Equations 13.4 and 13.11, is clear. There are, likewise, some differences. The first difference of note is that the sum of all momentum terms is not zero because there is a force term involved. The second difference, as mentioned earlier, is the vector character of Equation 14.2 in contrast to both Equations 13.4 and 13.11, which are scalar. For a rectangular coordinate system, Equation 14.2 can be written in component form as Equations 13.4 and 13.11

Fx =



Vˆ xe m ˙e −

e



Fy =

e

i

Vˆ ye m ˙e −

i

d Vˆ xi m ˙ i + Vˆ x,cv mcv dt

(14.3a)

d Vˆ yi m ˙ i + Vˆ y,cv mcv dt

(14.3b)

Newton’s Second Law of Motion

443

Section 1

Section 2 Atmosphere

r x

Flow Control Volume P1 = 530 kPa (abs)

d2 = 25 mm

d1 = 50 mm FIGURE 14.2 Water flowing through a horizontal nozzle in Example 14.1.

and



Fz =



Vˆ ze m ˙e −

e

i

d Vˆ zi m ˙ i + Vˆ z,cv mcv dt

(14.3c)

The utility of these expressions will become more obvious as we consider several examples.

14.3

Applications of the Control Volume Expression

Two critical choices must be made when applying the momentum theorem to a specific case. The choices pertain to the control volume and to the coordinate system. Various combinations will be illustrated in the examples to follow.

Example 14.1 Water flows through the nozzle shown in Figure 14.2, discharging to the atmosphere. The flow rate is 53 m3 /h with the other dimensions and conditions shown. Determine the force, including its direction, exerted on the nozzle by the coupling used to hold it in place.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. In this case, the choices of coordinate system and control volume are straightforward. Because the nozzle is stationary, we choose a coordinate system that is fixed with the x-axis oriented along the direction of flow. Moreover, because we are interested in the interaction between the nozzle and the coupling, we will choose the control volume that cuts through the point of contact, section 1, shown as a dashed line in Figure 14.2. The x-directional scalar form of the control volume applies so that the basic equation to be solved is Equation 14.3a

Fx =

e

Vˆ xe m ˙e −

i

d Vˆ xi m ˙ i + Vˆ x,cv mcv dt

(14.3a)

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Introduction to Thermal and Fluid Engineering

Moreover, because we are dealing with a steady-flow process, the accumulation term may be suppressed, leaving Fx = Vˆ xe m ˙e − Vˆ xi m ˙i e

i

The force may be expressed as

Fx = Fx + P1g A1

where the term Fx is the force exerted on the control volume at the location where the control surface interacts with its surroundings. In this case, Fx is exerted on the control volume by the coupling and is the unknown that we are seeking. Had the problem asked for the force that the nozzle exerts on the coupling, we would be looking for the reaction to Fx , namely Rx = −Fx . The unknown, Fx , is assumed to have its positive sense in the positive x direction. The solution will then indicate whether it is oriented in this manner; a positive result for Fx will mean that the resulting force on the control volume is to the right and a negative result will mean that it is to the left. The lone pressure-force term, P1g A1 , is the resulting x-directed pressure force because atmospheric pressure, exerted over the entire control surface, cancels. P1g is, thus, the net pressure and it is exerted on the control surface at section 1. This is in the positive x direction and this is the reason for its positive sign. We now examine the summation terms on the right-hand side of our equation. For this case, mass crosses the control surface only at sections 1 and 2. Thus, we can write Vˆ xe m ˙e − Vˆ xi m ˙ i = −m ˙ 1 Vˆ x1 + m ˙ 2 Vˆ x2 e

i

Because the flow is in at section 1, the sign is negative and the positive sign indicates that the flow is out at section 2. Both Vˆ x1 and Vˆ x2 are to the right and have positive signs. A further simplification is possible when, from a mass balance, we note that m ˙1 = m ˙ 2. The subscripts can then be dropped and the momentum flux terms become Vˆ xe m ˙e − Vˆ xi m ˙ i = m( ˙ Vˆ x2 − Vˆ x1 ) e

i

Thus, the complete momentum equation reduces to Fx + P1g A1 = m( ˙ Vˆ x2 − Vˆ x1 ) and the unknown, Fx , may be expressed as Fx = m( ˙ Vˆ x2 − Vˆ x1 ) − P1g A1 The velocities Vˆ x1 and Vˆ x2 can be expressed as ˙ V (53 m3 /h)(1 h/3600 s) Vˆ x1 = = = 7.5 m/s A1 (/4)(0.050 m) 2 and Vˆ x2 = 4Vˆ x1 = 4(7.5 m/s) = 30 m/s

Newton’s Second Law of Motion

445

We are now able to solve for Fx . With the density of water taken as 1000 kg/m3 and atmospheric pressure taken as Patm = 101 kPa, we have Fx = m( ˙ Vˆ x2 − Vˆ x1 ) − P1g A1 =

(53 m3 /h)(1000 kg/m3 )(30 m/s − 7.5 m/s) (3600 s/h)(1 kg-m/N-s2 )

 −[(530 − 101) × 103 N/m2 ] (0.050 m) 2 4

= 331 N − 842 N = −511 N ⇐ The force is negative, that is, it is acting toward the left.

Example 14.2 Consider the nozzle of Example 14.1 with all conditions the same except

that the nozzle discharge is inclined at an angle, , relative to the x direction. Determine (a) the force exerted on the nozzle by the coupling for the general case shown and (b) the values of Fx for  = 90◦ and  = 180◦ .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. (a) The control volume and the coordinate system are indicated in Figure 14.3. For this case, the total force exerted by the coupling, F, will have components in both the x- and y-coordinate directions. We must consider the two scalar component expressions Fx = ˙e − ˙i Vˆ xe m Vˆ xi m e

and



Fy =



i

Vˆ ye m ˙e −

e



Vˆ yi m ˙i

i

where, because this is a steady-flow case, the time-dependent terms have been suppressed. Section 1 y x

Section 2 Atmosphere θ

P1 = 520 kPa (abs) d1 = 50 mm FIGURE 14.3 Water flowing through a reducing pipe bend in Example 14.2.

d2 = 25 mm

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Introduction to Thermal and Fluid Engineering

Considering the first of these expressions, in the x direction, we have Fx + P1g A1 = m( ˙ Vˆ x2 − Vˆ x1 ) and for the y direction, we may write Fy = m ˙ Vˆ y2 Note that the atmospheric pressure exerted over the entire surface cancels leaving P1g A1 at section 1 as the only x directed net pressure force. Thus, it appears only in the x-component expression. One should also observe that the momentum influx, due to mass flow, occurs only at section 1 and, because Vˆ 1 is entirely x directed, this term also appears only in the x-component direction. We may now complete the solution for the total force. The operational equations for our two unknown force components are Fx = m( ˙ Vˆ x2 cos  − Vˆ x1 ) − P1g A1 and Fy = m ˙ Vˆ 2 sin  Using values already computed in Example 14.1, we obtain for Fx Fx = m( ˙ Vˆ x2 cos  − Vˆ x1 ) − P1g A1 =

(53 m3 /h)(1000 kg/m3 )(30 cos  m/s − 7.5 m/s) (3600 s/h)(1 kg-m/N-s2 )

 −[(530 − 101) × 103 N/m2 ] (0.050 m) 2 4

= (110.4 N)(4 cos  − 1) N − 842.4 N = 441.6 cos  N − 952.8 N and for F y Fy = m ˙ Vˆ y2 sin  =

(53 m3 /h)(1000 kg/m3 )(30 sin  m/s) (3600 s/h)(1 kg-m/N-s2 )

= 441.6 sin  N The force on the nozzle by the coupling is the vector sum of these components, F = (441.6 cos  − 952.8)x N + 441.6 sin y N ⇐ where x and y are unit vectors in the x and y directions. We may note that when the nozzle is horizontal,  = 0◦ , this expression yields a force in the x direction of F = −511.2x N which checks with the solution for Example 14.1

Newton’s Second Law of Motion

447

Nozzle Jet

1 Blade

Control Volume

y 2

135° x

FIGURE 14.4 Free-water jet striking a turning vane for Example 14.3.

(b) We can use our solution for part (a) F = (441.6 cos  − 952.8)x N + 441.6 sin y N to determine the force for any angle. If  = 90, ◦ cos  = 0, and sin  = 1.0 so that F = −952.8x N + 441.6y N ⇐ with a magnitude of F = 1050.2 N and if  = 180◦ , cos  = −1, and sin  = 0 F = −1394 N ⇐ Most of us have observed, firsthand, the behavior of a hose with water being discharged through a nozzle. Some thought about our findings in Examples 14.1 and 14.2 should reconcile our experiences relative to the results obtained. Our next example examines the effect of a free jet, that is one that issues from a nozzle and is not constrained by a tube wall, as it strikes an object that deflects the stream.

Example 14.3 A free jet of water flowing at a rate of 0.06 m3 /s and a velocity of 7.6 m/s strikes a fixed blade that turns the flow as indicated in Figure 14.4. Neglecting friction between the blade and the water jet (a) evaluate the force exerted by the blade on the water jet and (b) determine the force exerted on the jet if the blade moves to the right at a steady velocity of 3 m/s.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. (4) Negligible friction is specified. (a) We first specify the coordinate system to be fixed, with the x and y directions selected as indicated in Figure 14.4. The control volume, enclosed in the dashed line, is conveniently

448

Introduction to Thermal and Fluid Engineering

chosen so that the entering and leaving liquid streams are normal to the boundary. It is important to realize that the control volume will also cut through the blade support and that the net force exerted on the blade by the surroundings is the force exerted by the blade support. The x- and y-scalar forms of the momentum theorem, for steady flow, may be written as Fx = Vˆ xe m ˙e − Vˆ xi m ˙ i = m( ˙ Vˆ x2 − Vˆ x1 ) e

i ◦

ˆ = −m ˆ = m(− ˙ Vˆ cos 45 − V) ˙ V(cos 45◦ + 1) = −1.707m ˙ Vˆ and Fy =



Vˆ ye m ˙e −

e



Vˆ yi m ˙i

i

= m( ˙ Vˆ y2 − Vˆ y1 ) = −m ˙ Vˆ sin 45◦ = −0.707Vˆ With the density of water taken as 1000 kg/m3 , insertion of the proper numerical values allows the solution to be completed. Fx = −

(0.06 m3 /s)(1000 kg/m3 )(7.6 m/s)(1.707) 1 kg-m/N-s2

= −778.4 N

and Fy = −

(0.06 m3 /s)(1000 kg/m3 )(7.6 m/s)(0.707) 1 kg-m/N-s2

= −322.4 N

These are the components of the force exerted by the surroundings to hold the blade in place under the influence of the water jet. Because the problem statement specifies that we find the force exerted by the water on the blade, the correct answer to our problem is the negative of the force, F. Hence, the result is F = 778.4 x N + 322.4y N ⇐ (b) When the blade is moving to the right, the choice of the coordinate system and control volume is less obvious. To illustrate the effect of these choices, we will work the problem twice. First, we take the control volume as shown in Figure 14.4, that is, it moves with the blade and always has the same orientation relative to the blade. We will also choose a coordinate system that is fixed to the blade and is thus moving to the right at a fixed velocity of 3 m/s. Having made these choices, we can write the expressions for Fx and F y as Fx = Vˆ xe m ˙e − Vˆ xi m ˙ i = m( ˙ Vˆ x2 − Vˆ x1 ) e

i

and Fy =

e

Vˆ ye m ˙e −



Vˆ yi m ˙ i = m( ˙ Vˆ y2 − Vˆ y1 )

i

ˆ however, are different which are the same expressions as in part (a). The quantities m ˙ and V, in this case. The mass flow rate, m, ˙ across the boundary is equal to AVˆ r where Vˆ r is the

Newton’s Second Law of Motion

449

velocity relative the boundary. With the density of water taken as  = 1000 kg/m3 m ˙ = AVˆ r ˙ Vˆ r and with A = V/ m ˙ = (1000 kg/m3 )

0.06 m3 /s (7.6 m/s − 3.0 m/s) = 36.3 kg/s 7.6 m/s

The velocities, Vˆ x1 , Vˆ x2 , Vˆ y1 , and Vˆ y2 are all for the fluid stream crossing the boundary relative to the selected coordinate system. In our case, we have Vˆ x1 = 7.6 m/s − 3.0 m/s = 4.6 m/s Vˆ x2 = −4.6 cos 45◦ m/s Vˆ y1 = 0 Vˆ y2 = −4.6 sin 45◦ m/s We may now solve for Fx and F y Fx = m( ˙ Vˆ x2 − Vˆ x1 ) = =

(36.3 kg/s)(−4.6 m/s)(cos 45◦ + 1) 1 kg-m/N-s2 (36.3 kg/s)(−4.6 m/s)(1.707) 1 kg-m/N-s2

= −285.0 N and F y = m( ˙ Vˆ y2 − Vˆ y1 ) = =

(36.3 kg/s)(−4.6 m/s)(sin 45◦ ) 1 kg-m/N-s2 (36.3 kg/s)(−4.6 m/s)(0.707) 1 kg-m/N-s2

pjwstk|402064|1435600721

= −118.1 N These are the components of the force exerted by the surroundings as discussed earlier. The force exerted by the water on the blade will be F = 285.0x N + 118.1 y N ⇐ As a second approach to this problem, we will use the same moving control volume but choose a coordinate system fixed in space. This would be the case of a distant stationary observer watching as the blade passes by. The same equations for Fx and F y apply, that is Fx = m( ˙ Vˆ x2 − Vˆ x1 )

450

Introduction to Thermal and Fluid Engineering

and F y = m( ˙ Vˆ y2 − Vˆ y1 ) Because the control volume is still attached to the moving blade, the mass flow rate, m, ˙ will still be 36.3 kg/s. However, the velocity components will differ and will now be Vˆ x1 = 7.6 m/s Vˆ x2 = [(7.6 − 3)(− cos 45◦ ) m/s)] + 3 m/s Vˆ y1 = 0 Vˆ y2 = (7.6 − 3)(− sin 45◦ ) m/s Here, Vˆ x1 and Vˆ y1 , the velocity components at section 1 are straightforward; the velocities at this point are those of the jet. At section 2, the relationships are a bit more complex. The foregoing values of Vˆ x1 and Vˆ y1 are the vector sums of the velocity of the water relative to the blade plus the velocity of the blade relative to the fixed coordinate system. When we substitute into the expressions for Fx and F y , we obtain Fx = =

(36.3 kg/s){[(7.6 m/s − 3 m/s)(− cos 45◦ )] + 3 m/s − 7.6 m/s} 1 kg-m/N-s2 (36.3 kg/s)(7.6 m/s − 3 m/s)(− cos 45◦ − 1) 1 kg-m/N-s2

= −285.0 N

and Fy =

(36.3 kg/s)[(7.6 m/s − 3 m/s)(− sin 45◦ ) − 0 m/s] 1 kg-m/N-s2

= −118.1 N

Thus, the force exerted by the water on the blade will be F = 285x N + 118.1y N ⇐ It is reassuring to note that the same result was achieved even though the problem starting conditions were posed differently. The reader will likely prefer one approach over the other but proper analysis of any choice will yield a correct solution as long as all of the terms are evaluated correctly. In our next example, we will consider a case where the accumulation term is nonzero.

Example 14.4 A tank car with an open top is shown in Figure 14.5 at a time when it is in the path of a jet of water. The jet issues from a 1-cm-diameter nozzle and has a velocity of 15 m/s. The tank car moves to the right at 5 m/s. Determine the force exerted on the tank car by the water jet.

Newton’s Second Law of Motion

451

Water Jet 45°

1

y

Vj = 15 m/s x Vt = 5 m/s

Control Volume

FIGURE 14.5 Tank car receiving water from a jet in Example 14.4.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. The control volume, which is moving to the right with the tank car, is shown in Figure 14.5 by the dashed lines. We will first choose a coordinate system that also moves with the tank car. The x- and y-component forms of the momentum theorem are given by Equations 14.3a and 14.3b d Fx = Vˆ xe m ˙e − Vˆ xi m ˙ i + Vˆ x,cv mcv (14.3a) dt e i and

Fy =

e

Vˆ ye m ˙e −

i

d Vˆ yi m ˙ i + Vˆ y,cv mcv dt

(14.3b)

We first consider the x direction where Equation 14.3a reduces to d Fx = −Vˆ x1 (A1 V1 ) + Vˆ x, t m ˙ cv = −m( ˙ Vˆ j cos 45◦ − Vˆ t ) − m(0) ˙ dt In this expression, we have noted that, by mass conservation, the rate of mass accumulation in the tank is equal to the rate at which mass enters across the top boundary. The velocity of this fluid in the tank is zero relative to the coordinate system, which also moves with velocity Vˆ t . The x-component velocity relative to the moving coordinate system is seen to be Vˆ j cos ◦ 45 − Vˆ t . Solving for Fx , with the density of water taken as 1000 kg/m3 yields

 (1000 kg/m3 ) (0.15 m) 2 (15 m/s)(15 cos 45◦ m/s − 5 m/s) 4 Fx = − 1 kg-m/N-s2 = −1.486 kN

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Introduction to Thermal and Fluid Engineering

For F y , which is less complicated, we obtain F y = −Vˆ y1 m ˙1+

d ˆ V y, t m ˙ cv dt

= −m(− ˙ Vˆ j sin 45◦ ) + m(0) ˙

 1000 kg/m3 ) (0.15 m) 2 (15 m/s)(−15 sin 45◦ m/s) 4 =− 1 kg-m/N-s2 = 2.811 kN The force exerted by the water on the tank car has components that are of the opposite sign. Hence, the desired solution for the water force is F w = 1.486x kN + 2.811y kN ⇐ A stationary coordinate system will be chosen for an alternate solution. In this case, we may write for Fx and F y  d ˆ Fx = −Vˆ x1 m ˙1+ V x,cv m ˙ cv dt = −m( ˙ Vˆ j cos 45◦ ) + m ˙ Vˆ t = −m( ˙ Vˆ j cos 45◦ − Vˆ t ) and F y = −Vˆ y1 m ˙1+

 d ˆ V y,cv m ˙ cv = −m(− ˙ Vˆ j sin 45◦ ) + m(0) ˙ dt

These two solutions, although developed differently, are identical to those obtained in the previous approach for a moving coordinate system. We further illustrate the momentum theorem in Design Example 7 that now follows.

14.4

Design Example 7

A concrete slurry mixture with a density of 2500 kg/m3 is dumped from a hopper onto a conveyer belt as shown in Figure 14.6. The slurry is delivered to the conveyer at a rate of 600 m3 /h. For conveyer drive wheels of 50-cm diameter rotating at 300 rpm, estimate the power required to drive the conveyer.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The concrete slurry is incompressible. (3) Uniform properties exist at each cross section considered. A stationary control volume is chosen as shown in Figure 14.7 and we may apply the directional component form of the momentum equation to this stationary control volume. Fx = Vˆ xe m ˙e − Vˆ xi m ˙i

Newton’s Second Law of Motion

453

FIGURE 14.6 Conveyor belt for Design Example 7.

and for this single input-single output situation, we have m ˙e =m ˙i =m ˙ and Fx = m( ˙ Vˆ xe − Vˆ xi ) We note that the slurry entering the control volume at point 1 has no horizontal velocity component. Thus, Vˆ xe = Vˆ x and Fx = m ˙ Vˆ x ˙ = Fx Vˆ x , we have Because the required power is W ˙ = Fx Vˆ x = ( m W ˙ Vˆ x ) Vˆ x = m( ˙ Vˆ x ) 2 The velocity Vˆ x is related to the velocity of the drive wheels Vˆ x = r  = (0.25 m)(2 rad/rev)(300 rev/min)

1 min/s = 7.85 m/s 60

and with m ˙ = (600 m3 /h)(2500 kg/m3 )(1 h/3600 s) = 416.7 kg/s

1

Control Volume

2

FIGURE 14.7 Stationery control volume for Design Example 7.

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we have ˙ = m( W ˙ Vˆ x ) 2 = (416.7 kg/s)(7.85 m/s) 2 = 25,700 W

14.5

(25.7 kW) ⇐

The Control Volume Relation for the Moment of Momentum

The treatment of angular momentum or moment of momentum is a direct extension of that just considered for linear momentum. Operationally, we will produce a moment by forming the vector or cross product of a displacement vector, r, with each term in our fundamental relationship, Equation 14.2, r×



F=r×



Vˆ e m ˙e −r×

e

i

d Vˆ i m ˙ i + r × Vˆ cv mcv dt

(14.4)

An evaluation of the left-hand side produces r× F= r×F = M where M is the net external moment exerted on the control volume by its interaction with the surroundings. The terms on the right-hand side become r×



Vˆ e m ˙e −r×

e

i

=



d Vˆ i m ˙ i + r × Vˆ cv mcv dt

(r × Vˆ e ) m ˙e −

e

d (r × Vˆ i ) m ˙ i + (r × Vˆ cv )mcv dt i

With these modifications, Equation 14.4 may be written in operational form as

M=

e

(r × Vˆ e ) m ˙i −

i

d (r × Vˆ i ) m ˙ i + (r × Vˆ cv )mcv dt

(14.5)

This expression is fundamental for the application of Newton’s second law to a rotating system. In Cartesian coordinates, the component forms of Equation 14.5 are       d Mx = (r × Vˆ e )  m ˙e − (r × Vˆ i )  m ˙ i + (r × Vˆ cv )  mcv (14.6a) dt x x x e i

      d   My = (r × Vˆ e )  m ˙e − (r × Vˆ i )  m ˙ i + (r × Vˆ cv )  mcv dt y y y e i

(14.6b)



      d   ˆ ˆ ˆ Mz = (r × V e )  m ˙e − (r × V i )  m ˙ i + (r × V cv )  mcv dt z z z e i

(14.6c)

It is worth noting some characteristics of the vector operations that are indicated in Equations 14.5 and 14.6. First, one must be careful to observe that the vector cross product is not commutative, that, is r × Vˆ =  Vˆ × r. For these equations to be valid, care should be ˆ taken to always form the cross product as r × V.

Newton’s Second Law of Motion

455 F

rx

r ry

y x z FIGURE 14.8 Vector representation for a moment.

We may also recall that a convenient means for evaluating a cross product is by expanding a 3 × 3 determinant as    x y z       r × V =  r x r y r z     Vˆ x Vˆ y Vˆ z  = (r y Vˆ z − r z Vˆ y )x + (r z Vˆ x − r x Vˆ z )y + (r x Vˆ y − r y Vˆ x )z and we take note of the components of r × V   (r × V)  = r y Vˆ z − r z Vˆ y x

  (r × V)  = r z Vˆ x − r x Vˆ z y

  (r × V)  = r x Vˆ y − r y Vˆ x z

The result of the vector product, r × F, is a vector whose direction is perpendicular to the plane of the two vectors, r and F. We may recall, from solid mechanics, that a force, F, applied to a socket wrench, as shown in Figure 14.8, will have a vector character that identifies the axis about which the turning tendency is induced. If, in Figure 14.8, the force, F, is in the z direction, then the resulting moment is evaluated as   x y z     M = r × F = r x r y 0  = r y F x − r x F y   0 0 F

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Introduction to Thermal and Fluid Engineering

In this case, we see that the moment, M, is normal to the plane of the two vectors, r and F. It is also clear that if r y = 0, r × F would equal −(F × r), that is, the vector M will point in the −y direction. These ideas and conventions will become clearer as we consider some examples of moment of momentum applications.

14.6

Applications of the Moment of Momentum Relationship

Example 14.5 Consider Figure 14.9 in which the free-water jet is issuing from a nozzle and striking a fixed blade as was analyzed in Example 14.3. In this example, we want to consider that the blade is connected to a shaft at some distance, r , which is normal to the plane of Figure 14.9. The hub of this blade is oriented such that all actual or induced rotary motion will be about the y axis. The displacement vector, r, would then be equal to r z. For this case, evaluate the moment about the y axis produced by the water jet as it strikes the blade and has its direction changed. Evaluate the following cases: (a) the blade is stationary and (b) the blade possesses a tangential velocity of 3 m/s at the point of application of the water jet. The radial distance from the axis of rotation is 12.5 cm. As in Example 14.3, the jet exits the nozzle with m ˙ = 0.06 kg/s and Vˆ = 7.6 m/s.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. The control volume choice is indicated by the dashed lines in Figure 14.9. The control volume extends in the z direction far enough to include the axis of rotation and cuts through the shaft that is oriented with its axis in the y direction. We will consider a coordinate system that is fixed on this axis, with x and y pointing radially outward. The starting point for the solution of this problem is Equation 14.6b. (a) Because steady flow exists, Equation 14.6b with the accumulation term suppressed is     My = (r × V e )  m ˙e − (r × V i )  m ˙i (14.6b) y

e

Nozzle

y

i

Blade Jet y

Control Volume

x 135°

FIGURE 14.9 Free-water jet striking a turning vane for Example 14.5.

Newton’s Second Law of Motion

457

Then, observing that mass crosses the boundary at sections 1 and 2, we have My = (−r Vˆ j cos 45◦ m) ˙ − (r Vˆ j m) ˙ = −r Vˆ j m(1 ˙ + cos 45◦ ) With appropriate numerical values substituted, we obtain My = −

(0.125 m)(7.6 m/s)(1000 kg/m3 )(0.06 m3 /s)(1.707) 1 kg-m/n-s2

= −97.3 N-m

This is the moment exerted by the surroundings (the shaft in this case) to maintain the blade stationary under the influence of the jet. The quantity that we seek is the moment produced by the water jet, which is the negative of My . Thus, My,water = 97.3 N-m ⇐ (b) The same coordinate system and control volume employed in (a) will be used here. The expression that applies, obtained from Equation 14.6b, is     My = r Vˆ x  m ˙ − r Vˆ x  m ˙ 2

1

where, because of the motion of the control volume, m, ˙ Vˆ x1 and Vˆ x2 will have values different than those used in (a). The proper values for these quantities are ˆ in = (AV) ˆ out m ˙ = (AV) Vˆ x1 = 7.6 m/s and Vˆ x2 = (7.6 m/s − 3.0 m/s)(− cos 45◦ ) + 3.0 m/s = −3.25 m/s + 3.0 m/s = −0.25 m/s Having these values, we may complete the solution to obtain My = mr ˙ ( Vˆ x2 − Vˆ x1 ) =

(0.125 m)(1000 kg/m3 )(0.06 m3 /s)(−0.25 m/s − 7.6 m/s) 1 kg-m/n-s2

= −58.87 N-m As before, the moment produced on the shaft by the water is My,water = 58.87 N-m ⇐ In the foregoing example, we have performed the basic analysis of a water turbine. In the case of a real turbine, there would, naturally, be many blades attached to a rotating hub, where as one blade rotates out of the path of the water jet, another rotates into position, so that the process is continued.

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Introduction to Thermal and Fluid Engineering

We might also note that maximum moment is produced when there is no rotation. However, the desired output from a turbine is power, which is the product of angular velocity and the moment (or torque) produced. Thus, we are interested in the combination of angular velocity and moment that will produce maximum power. A turbine is one type of rotating machine that is of interest to engineers. Chapter 18 discusses, in some detail, the concepts relating to rotating machinery and their operational features.

Example 14.6 A schematic of a lawn sprinkler is shown in Figure 14.10. Water flows through the sprinkler at a rate of 0.001 m3 /min and the sprinkler rotates at 40 rpm. Determine the torque due to friction at the sprinkler hub.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Uniform properties exist at each cross section considered. The control volume and coordinate system are shown in Figure 14.10. The sought-after moment will be z directed so the moment of momentum expression to be solved is Equation 14.6c       d   Mz = (r × V e )  m ˙e − (r × V i )  m ˙ i + (r × V cv )  mcv (14.6c) dt z z z e i where, for steady-state operation, the accumulation term will be zero, leaving      Mz = (r × V e )  m ˙e − (r × V i )  m ˙i z

e

z

i

In evaluating the remaining terms, we note that mass crosses the control surface at three locations; it enters vertically through the hub and exits from the two sprinkler heads. Accordingly, the expression for Mz reduces to Mz = (r × V 2 ) m ˙2− (r × V 1 ) m ˙1 2

1

where section 1 is the entering section and section 2 represents the surface across which the flow leaves. Discharge Jet Diameter = 5 mm x V z 2 V

15 cm

35° V =. 0.001 m3/min FIGURE 14.10 Lawn sprinkler for Example 14.6.

1

ω = 40 rpm

Newton’s Second Law of Motion

459

Vj

Vj

z

30°

ω

FIGURE 14.11 View of the lawn sprinkler of Example 14.6 looking down at the radial part of the sprinkler arm.

Because the coordinate reference frame is stationary, we must think through the representation of the exiting velocity. Figure 14.11 examines the exit flow from a perspective looking down the radial part of the sprinkler arm. With the cross-sectional area of the jet openings expressed as

 Aj = 2 (0.005 m) 2 = 3.927 × 10−5 m2 4 the magnitude of the jet velocity relative to the sprinkler arm is seen to be ˙ V (0.001 m3 /min)(1 min/60 s) Vˆ j = = = 0.424 m/s Aj 3.927 × 10−5 m2 The tangential velocity of the sprinkler arm is Vˆ t = r  = (0.15 m)(40 rev/min)(2 rad/rev)(1 min/60 s) = 0.628 m/s The velocity of one exit stream, relative to a fixed coordinate frame, is the vector sum of the jet and tangential velocities. The absolute velocity can thus be expressed as V 2 = (−Vˆ j cos 30◦ θ + Vˆ j sin 30◦ z + Vˆ t θ) m/s = [(−0.424 cos 30◦ + 0.628)θ + 0.424 sin 30◦ z] m/s = 0.261θ m/s + 0.212z m/s We may also represent the inlet velocity as V 1 = Vˆ 1 z and the solution for M, a vector, is now a matter of substituting the foregoing quantities into M = 2(r × V 2 ) m ˙ − (r × V 1 ) m ˙ or   r   M = 2 −0.15   0

  r z     0 0 m ˙ − 0   0 0.261 0.212 θ

θ 0 0

 z  0  (−m) ˙   ˆ V1

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Introduction to Thermal and Fluid Engineering

or M = (−0.15)(0.261)z N-m − (−0.15)0.212)θ N-m We may now obtain a numerical value for the component of M in the z-coordinate direction as Mz =

2(−0.15 m)(0.001 m3 /60 s)(0.261 m/s)(1000 kg/m3 ) 1 kg-m/N-s2

= −0.00131 N-m ⇐ This is the result sought, the torque imposed on the sprinkler shaft by friction. This torque balances the moment of momentum produced by the water jets. Notice that the external moment exerted on the shaft is negative.

14.7

Summary

In this chapter we have developed and applied the control volume expression for Newton’s second law of motion. The two fundamental integral expressions are Equation 14.2 for linear momentum d F= Ve m ˙e − Vi m ˙ i + V cv mcv (14.2) dt e i and Equation 14.5 for the moment of momentum d M= (r × V e ) m ˙e − (r × V i ) m ˙ i + (r × V cv )mcv dt e i

(14.5)

These expressions are extremely powerful and form the basis for a significant portion of fluid flow analysis.

14.8

Problems

The Cross Product 14.1: Form the cross product A × B if A = 2x + 3y + 4z and

B = −x + 2z

14.2: Form the cross product A × B if A = x + 2y + 4z

and

B = x + 4z

14.3: Form the cross product A × B if A = 2y and

B = −8z

14.4: Form the cross product A × B if A=x+y+z

and B = 2(x − y − z)

14.5: Form the cross product A × B if A = x + 2y − 3z

and B = −y + 2z

Newton’s Second Law of Motion

461

14.6: Form the cross product A × B if A = x − 2y − 3z

and B = x + y + 2z

14.7: Form the cross product A × B if A = 2x + y

and B = x − 2z

14.8: Form the cross product A × B if A = 2z

and B = −8x

14.9: Form the cross product A × B if A = 24x − 36y + 48y

and B = 12x − 24y

14.10: Form the cross product A × B if A = 6x − 8y + 10z

and

B = 2(3x + 4y − 5z)

Forces Developed by Fluids in Motion 14.11: Mud, at a rate of 3200 kg/s, is discharged to a barge at a velocity of 1.25 m/s as shown in Figure P14.11. Determine the tension on the mooring line.

30°

FIGURE P14.11

14.12: Water is pumped through a 30-cm-diameter pipe entering the bow of a boat at Vˆ = 2.05 m/s. The pipe tapers to a diameter of 20 cm at the stern of the boat where it is discharged. Determine the thrust developed. 14.13: In Figure P14.13, the vane manages to turn the jet completely around. If the maximum support force is Fo , determine an expression for the maximum water jet velocity, Vˆ o .

Fo

ρo, do, Vo

FIGURE P14.13

14.14: In the sluice gate arrangement shown in Figure P14.14, uniform flow and hydrostatic pressure may be assumed at section 1 upstream of the gate and section 2 downstream of the gate. Derive an expression for the force, F, required to hold the ˆ g, h 1 , and h 2 . gate in place as a function of , V,

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Introduction to Thermal and Fluid Engineering 1

h1

h2

V1 V 2

2 FIGURE P14.14

14.15: A small box rests on a platform which, in turn, rests on a steady water jet with a flow area of 0.03 m2 . If the total weight is 750 N, determine the jet velocity. 14.16: The tank in Figure P14.16 has an empty weight of 925 N and holds 1.125m3 of water at 20◦ C. The entrance and exit pipes both have a 6-cm diameter and both carry 82.5 L/min. Determine the value of the scale reading, W. 6 cm 82.5 L/min

W Water 6 cm

FIGURE P14.16

14.17: A 2.5-cm-diameter jet of water hits a flat plate held normal to the axis of the jet. The force exerted by the jet on the plate is 650 N. Determine the volumetric flow rate of the water. 14.18: A jet of water having a velocity of 12.5 m/s and a volumetric flow rate of 6×10−4 m3 /s is deflected through a right angle as shown in Figure P14.18. Neglect friction and determine the magnitude and direction of the resultant force.

90°

Water Jet

FIGURE P14.18

14.19: In Figure P14.19, the spring has a stiffness of 1.625 kN/m. Determine the force on the wheels caused by the deflection of the water jet and the compression of the spring.

Newton’s Second Law of Motion

463 d = 37.5 cm

1.625 kN/m 143.13° d = 37.5 cm V = 18 m/s Spring

FIGURE P14.19

14.20: The water tank in Figure P14.20 sits on a frictionless cart and feeds a jet of of 7.5 cm in diameter at a velocity 12 m/s. Determine the tension in the cable. 6m

45°

Cable

FIGURE P14.20

14.21: A free jet of water is deflected by a 90◦ curved vane as indicated in Figure P14.21. The water jet has a diameter of 4 cm and a velocity of 8 m/s. Determine the force required to hold the vane in place.

90°

FIGURE P14.21

14.22: Refer once again to Problem 14.14 and the sluice gate in Figure P14.14. If Vˆ 1 = 1.375 m/s, h 1 = 4.125 m, and h 2 = 0.75 m, determine the force per unit width required to hold the gate in place. 14.23: The water jet in Figure P14.23 moving at 15 m/s strikes a stationary splitting vane such that 40% of the water moves downward. Assume ideal flow in the horizontal plane and determine the magnitude and direction of the force on the splitter.

. V = 0.12 m3/s 53.13°

FIGURE P14.23

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14.24: The horizontal nozzle in Figure P14.24 has diameters of d1 = 25 cm and d2 = 15 cm. The inlet pressure is 400 kPa and the exit velocity is 27.5 m/s. Assuming incompressible steady flow, determine the total tensile force required to hold the nozzle stationary.

Water d1 = 25 cm P = 400 kPa

d2 = 15 cm

V = 27.5 m/s

2

1

FIGURE P14.24

14.25: A hose and nozzle discharges a horizontal water jet against a vertical plate as indicated in Figure P14.25. The volumetric flow rate of the water is 0.045 m3 /s and the diameter of the nozzle is 5 cm. Determine the force necessary to hold the plate in place. . Jet, V = 0.045 m3/s

5 cm FIGURE P14.25

Linear Momentum 14.26: A two-dimensional object is placed in a 1.6-m-wide water tunnel as shown in Figure P14.26. The upstream velocity, Vˆ 1 , is uniform across the cross section. For the downstream velocity profile shown, determine the value of Vˆ 2 .

1.60 m 0.60 cm

V1 = 6.15 m/s

V2 FIGURE P14.26

14.27: A water jet pump has an area, Aj = 7.50 × 10−3 m2 and a jet velocity of Vˆ j = 17.5 m/s, which entrains a second stream of water having a velocity of Vˆ s = 3.2 m/s in a constant area pipe of total area A = 0.0560 m2 . The configuration is shown in Figure P14.27 and, at section 2, the water is throughly mixed. Assuming onedimensional flow and neglecting wall shear, determine (a) the average velocity of mixed flow at section 2 and (b) the pressure rise, P2 − P1 , assuming the pressure of the jet and secondary stream to be the same at section 1.

Newton’s Second Law of Motion

465 1

2 Vs

A1

Vj

V2

FIGURE P14.27

14.28: The pressure on the control volume and the x components of velocity are as illustrated in Figure P14.28. Assuming incompressible flow, determine the forces exerted on the cylinder by the field.

6d d

Vo

Vo FIGURE P14.28

14.29: The volumetric flow rate of a jet of water is 0.056 m3 /s and its velocity is 8 m/s. Determine (a) the magnitude of the x and y components of the force exerted on the fixed blade in Figure P14.29 and (b) the magnitude and velocity of the water jet leaving the blade if the blade is moving to the left at 4.8 m/s. 8 m/s

30°

FIGURE P14.29

14.30: A plate moves perpendicularly toward a discharging jet at 1.5 m/s. The jet discharges water at the rate of 0.056 m3 /s and a velocity of 9.6 m/s; the flow is frictionless. Determine (a) the force of the fluid on the plate and (b) the force of the fluid on the plate if the plate were stationary. 14.31: Air with a density of  = 1.2 kg/m3 flows in the 25-cm-diameter duct at 10 m/s. The duct is choked at its exit by the presence of a 90◦ cone as shown in Figure P14.31. Neglect air friction and estimate the force of the airflow on the cone.

466

Introduction to Thermal and Fluid Engineering 1 cm

25 cm

90°

40 cm

FIGURE P14.31

14.32: The water jet in Figure P14.32, is 12 cm in diameter with a volumetric flow rate of 0.10 m3 /s. It flows over a fixed cone with a base diameter of 40 cm. The water leaves as a conical sheet with the same velocity as the jet. Determine the force in newton required to hold the cone fixed.

30° F Base diameter, d

FIGURE P14.32

14.33: Oil with a specific gravity of 0.80 flows smoothly through the circular reducing section shown in Figure P14.33 at a volumetric flow rate of 0.085 m3 /s. The entering and leaving velocity profiles are uniform. Estimate the force that must be applied to the reducer to hold it in place. 1 2

ρ2g = 35 kPa D = 6.25 cm

ρ1g = 350 kPa D = 30 cm FIGURE P14.33

14.34: A nozzle that discharges a jet of water having a diameter of 3.60 cm into the atmosphere is at the end of a 7.5-cm-diameter pipe. The pressure in the pipe is 400 kPa gage and the rate of discharge is 1.50 m3 /min. Determine the magnitude and direction of the force required to hold the nozzle to the pipe. 14.35: Water flows steadily through the horizontal 30◦ pipe bend shown in Figure P14.35. At station 1, the diameter is 30 cm, the velocity is 12 m/s and the pressure is 128 kPa

Newton’s Second Law of Motion

467

gage. At station 2 the diameter is 38 cm and the pressure is 145 kPa gage. Determine the forces, Fx and F y , necessary to hold the pipe bend stationary.

x

30° z 2

1 FIGURE P14.35

14.36: The rocket nozzle shown in Figure P14.36 consists of three welded sections. Determine the axial stress at junctions 1 and 2 when the rocket is operating at sea level and the mass flow rate is 350 kg/s. 1

2

Welds

V = 275 m/s 1050 m/s d = 45 cm 30 cm P = 7 MPa 3.75 MPa

3

Thickness = 0.953 cm

2060 m/s 60 cm 180 kPa FIGURE P14.36

14.37: Figure P14.37 represents an open tank car that travels to the right at a uniform velocity of 4.5 m/s. At the instant shown, the car passes under a jet of water issuing from a 10-cm-diameter pipe at a velocity of 10 m/s. Determine the force exerted on the tank car by the water jet. 45°

4.5 m/s

FIGURE P14.37

14.38: The open tank car, shown in Figure P14.38, travels to the right at a velocity of Vˆ c m/s. A jet of area Aj exhausts the fluid of density  at a velocity of Vˆ j m/s relative to the tank car. The tank car, at the same time, collects from an overhead sprinkler fluid that directs fluid downward with velocity Vˆ g . Assume that the sprinkler flow is uniform

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Introduction to Thermal and Fluid Engineering over the car area. In terms of the quantities indicated and determine the net force of the fluid on the tank car.

V = 10 m/s Vj Aj

Vc = 4.50 m/s

L

FIGURE P14.38

14.39: The cart shown in Figure P14.39 is supported by wheels and a linear spring. The water jet is deflected 45◦ by the cart. Determine (a) the force on the wheels caused by the jet and (b) the spring deflection compared with its unstressed position. d2 = 3 cm 45° V1 = 25 m/s Water d1 = 3 cm

Linear spring 1500 N/m FIGURE P14.39

14.40: The test section of a stationary jet engine is shown in Figure P14.40. Air with density 1.275 kg/m3 enters as shown. The inlet and outlet cross-sectional areas are both 1 m2 and the mass of fuel consumed is 1% of the mass of the air entering the test section. For these conditions, determine the thrust developed by the engine.

V1 90 m/s

V2 275 m/s

Fuel FIGURE P14.40

14.41: For the system of Problem 14.26, the total drag on the object is measured to be 600 N/m of length normal to the direction of flow. Neglecting frictional forces, determine the pressure difference between the inlet and outlet conditions. 14.42: The pump in the boat shown in Figure P14.42 pumps 0.25 m3 /s through the submerged water passage that has an area of 0.030 m2 at the bow and 0.180 m2 at the stern. Assuming that the inlet and exit pressures are equal, determine the tension in the retraining rope.

Newton’s Second Law of Motion

469

30°

Pump FIGURE P14.42

14.43: The boat shown in Figure P14.43 is being driven at a steady speed, Vˆ o , by a jet of compressed air issuing from a 3-cm-diameter hole at Vˆ e = 350 m/s. Jet exit conditions are Pe = 100 kPa and Te = 20◦ C. The drag force on the hull as it moves through the water is given by K Vˆ 2o , where K = 20 N/(m/s)2 . The air drag on the upper surfaces of the boat is negligible. Determine the steady boat speed, Vˆ o .

de = 3 cm Ve

Compressed Air V0

 02 Hull Drag, kV FIGURE P14.43

14.44: Figure P14.44 considers a shock wave moving to the right at Vˆ o m/s. The properties ahead and behind the shock wave are not a function of time. By using the illustrated control volume, show that the pressure difference across the shock is P2 − P1 = 1 Vˆ w Vˆ 2

P2

P1 Vo

ρ2

ρ1

V2

V1 = 0

pjwstk|402064|1435600741

x

y FIGURE P14.44

14.45: If the shock-wave velocity in Problem 14.44 is approximately the speed of sound, determine the pressure change required to cause a velocity change of 3 m/s in (a) air at standard conditions ( Vˆ c = 344 m/s) and (b) water (Vˆ c = 1430 m/s). 14.46: Water enters section 1 in the pipe bend shown in Figure P14.46 at a volumetric flow rate of 0.20 m3 /s and exits at a 30◦ angle at section 2. With r as a radial coordinate and R as the pipe radius, section 1 has the laminar profile

r2 Vˆ = Vˆ m1 1 − 2 R

470

Introduction to Thermal and Fluid Engineering while section 2 has changed to a turbulent profile

r 1/7 Vˆ = Vˆ m2 1 − R For steady incompressible flow, what are the maximum velocities, Vˆ m1 and Vˆ m2 , in meters per second? 2 1

d2 = 6 cm

d1 = 10 cm 30°

FIGURE P14.46

14.47: The plate shown in Figure P14.47 is inclined at an angle of 53.13◦ and the flow is frictionless. Determine the forces Fx and F y necessary to maintain its position.

4 V = 30 m/s

3 Fx

•

m = 2 kg/s Fy

FIGURE P14.47

14.48: A steady, incompressible, frictionless, two-dimensional jet of fluid with breadth, h, ˆ and unit width impinges on a flat plate held at an angle, , to its axis as velocity, V, indicated in Figure P14.48. Neglecting gravity forces, determine (a) the total force on the plate and the breadths a and b of the two branches and (b) the distance, , to the center of pressure along the pipe from point 0. V Flowing fluid

h

ℓ a

0 α

b

FIGURE P14.48

14.49: Water flows in a pipe at 4 m/s. A valve at the downstream end of the pipe is suddenly closed. Determine the pressure rise in the pipe.

Newton’s Second Law of Motion

471

14.50: A liquid of density, , flows through a sluice gate as shown in Figure P14.50. The upstream and downstream flows are uniform and parallel so that the pressure variations at stations 1 and 2 may be considered to be hydrostatic. Determine (a) the velocity at station 2 and (b) the force per unit width R necessary to hold the sluice gate in place. P = Patm R h

V1

P = Patm L

1

V2

2 FIGURE P14.50

14.51: For the pipe flow reducing section shown in Figure P14.51, d1 = 8 cm, d2 = 5 cm, and P2 is approximately atmospheric. The entrance velocity Vˆ 1 is 6 m/s and the manometer reading is h = 28 cm. Estimate the force resisted by the flange bolts. 1

2 P2 ~ ~ Patm ~ ~ 100 kPa

Water

h Mercury FIGURE P14.51

14.52: A jet engine on a test stand (Figure P14.52) takes in atmospheric air at 20◦ C at section 1, where Vˆ 1 = 200 m/s and A1 = 0.30 m2 . The fuel to air ratio is 1:25 and the air leaves at section 2 where the area is A2 = 25 m2 at atmospheric pressure with a velocity of Vˆ 2 = 1000 m/s. Determine the test stand reaction that balances the thrust of the engine in steady flow, Rx . mfuel

1

Combustion Chamber

2

Rx FIGURE P14.52

14.53: In Figure P14.53, the water jet strikes normal to the plate. Neglect gravity and friction and determine the force, F , in newton required to hold the plate fixed.

472

Introduction to Thermal and Fluid Engineering Plate

Dj = 10 cm F Vj = 8 m/s

FIGURE P14.53

14.54: The steady-flow entrance case, shown in Figure P14.54, develops from uniform flow with velocity, Vˆ o , at section 1 to the laminar profile

r2 Vˆ = Vˆ max 1 − 2 R at section 2. Determine the wall drag as a function of P1 , P2 , , Vˆ o , and R. 2

r=R 1 r Vo

x

Friction Drag on Fluid FIGURE P14.54

14.55: A vane of turning angle,  = 120◦ , is mounted on a water tank as shown in Figure P14.55. It is struck by a 5-cm-diameter water jet flowing with a velocity of 16 m/s, which turns and falls into the water without spilling. Determine the force required to hold the tank stationary if  = 45◦ , 60◦ , and 90◦ . Vj = 16 m/s

dj = 5 cm

θ

Water

F

FIGURE P14.55

14.56: A dam discharges into a channel of constant width as shown in Figure P14.56. It is observed that water backs up behind the jet to a height, H. The velocity and height of the flow in the channel are given as Vˆ and h, respectively, and the density of the water is . Neglect the horizontal momentum of the flow that is entering the control

Newton’s Second Law of Motion

473

volume from above and assume friction to be negligible. Taking the air pressure in the cavity below the crest of falling water as atmospheric, use the momentum theorem and the control surface indicated to determine H.

H h

V

FIGURE P14.56

Moment of Momentum 14.57: The pipe shown in Figure P14.57 has a slit of thickness 0.0984 cm, so shaped that a sheet of water of uniform thickness, 0.0984 cm, issues out radially from the pipe. The velocity is constant along the pipe as shown and a volumetric flow rate of 0.90 m3 /s enters at the top. Determine the moment on the tube about the axis-B B. B

A = 2.581 × 10–3 m2

Slit

B 1m

2m Top View

FIGURE P14.57

14.58: Figure P14.58 shows a vane with a turning angle, , which moves with a steady speed ˆ (a) Assuming of Vˆ c . The vane receives a jet that leaves a fixed nozzle with speed, V. that the vane is mounted on rails as indicated in Figure P14.58, show that the power transmitted to the cart is Vˆ c /Vˆ = 1/3 and (b) assuming that there are a large number of such vanes mounted to a rotating wheel with a peripheral speed of Vˆ c , show that the power transmitted is maximum when Vˆ c /Vˆ = 1/2. θ

V Vc

FIGURE P14.58

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Introduction to Thermal and Fluid Engineering

14.59: A lawn sprinkler consists of two sections of curved pipe rotating about a vertical axis as shown in Figure P14.59. The sprinkler rotates with an angular velocity, , and the ˙ = 2Vˆ r A, effective discharge area is A. Thus, the water is discharged at the rate of V ˆ where Vr is the velocity of the water relative to the rotating pipe. A constant friction torque resists the motion of the sprinkler. Find an expression for the speed of the sprinkler in terms of the significant variables.

ω α

V r

R FIGURE P14.59

14.60: An air jet of diameter, d1 enters a series of moving blades as indicated in Figure P14.60 at absolute velocity, Vˆ 1 and angle 1 and leaves at absolute velocity, Vˆ 2 , and ˆ Assuming incompressible flow, if Vˆ 1 = angle, 2 . The blades move at speed, V. ◦ ˆ 60 m/s, V 2 = 40 m/s, 1 = 30 , and 2 = 60◦ , (a) find the velocity of the blades, Vˆ and (b) if  = 1.2 kg/m3 and d1 = 5 cm, and (c) find the power in watts applied to the blade for  = 1.2 kg/m3 and d1 = 5 cm.

α2 α1

β2 V2

V V1 β1 Air Jet

Blades

d1

FIGURE P14.60

14.61: Seawater,  = 1025 kg/m3 , flows through the impeller of the centrifugal pump shown in Figure P14.61 at a volumetric flow rate of 0.0520 m3 /s. Data and dimensions are n = 1180 rpm

1 = 5 cm

r1 = 5.00 cm

2 = 135◦

r2 = 20 cm

2 = 1.5 cm

Assuming that the absolute velocity of the water entering the impeller is radial, determine the torque exerted on the impeller by the fluid and the power required to drive the pump.

Newton’s Second Law of Motion

475

t1 t2

θ2

θ1

r2 r1

z

y

x ω

FIGURE P14.61

14.62: In Problem 14.61, determine (a) the angle 1 such that the entering flow is parallel to the vanes and (b) the axial load on the shaft for a shaft diameter of 2.5 cm and atmospheric pressure at the pump inlet. 14.63: A water sprinkler consists of two 1-cm-diameter jets at the ends of a rotating hollow rod as shown in Figure P14.63. The water leaves at 6 m/s. Determine the torque necessary to hold the sprinkler in place. 30°

12 in

FIGURE P14.63

14.64: Water enters the rotor shown in Figure P14.64 with a volumetric flow rate of 0.015 m3 /s, along the axis of rotation. The cross-sectional area of each of the three nozzles is 2 cm2 . Determine (a) the magnitude of the resisting torque necessary to hold the rotor stationary for  = 30◦ and (b) how fast the rotor will spin steadily if the resisting torque is reduced to zero and  = 0◦ , 30◦ , and 60.◦ Nozzle exit area normal in relative velocity = 2 cm2

θ

Top View 0.5 m

Side View

. V = 0.015 m3/s FIGURE P14.64

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Introduction to Thermal and Fluid Engineering

14.65: Front and side views of a centrifugal pump rotor or impeller are shown in Figure P14.65. The pump delivers 220 L/s of water and the blade exit angle is 35◦ from the tangential direction. The flow entering the rotor blade row is essentially radial as viewed from a stationary frame. Determine the power requirement. 35°

r1 = 9 cm

r2 = 15 cm 3000 rpm

3 cm FIGURE P14.65

Jets and Rockets 14.66: A rocket engine burns 5.5 kg of fuel and oxidizer per second and the combustion products leave the rocket at a velocity of 565 m/s relative to the rocket at the ambient air pressure. Determine the total thrust produced by the rocket engine. 14.67: A jet aircraft traveling at a rate of 200 m/s takes in air at the rate of 22.5 kg/s. The air fuel ratio is 25:1 and the exhaust velocity relative to the jet is 600 m/s. Determine the thrust produced by the engine. 14.68: A jet engine is being tested in the laboratory. The engine consumes 22.75 kg/s of air and 0.225 kg/s of fuel. The exit velocity of the gases is 475 m/s. Determine the thrust. 14.69: A jet engine under test uses 22.65 kg/s of air and 0.2265 kg/s of fuel. The exit velocity of the products of combustion is 460 m/s. Determine the thrust being developed by the engine. 14.70: A jet engine in an aircraft flying at 182.5 m/s uses 22.5 kg/s of air. Determine the velocity of the combustion products in order to develop a thrust of 6720 N. 14.71: The mass flow rate of the products of combustion for a rocket engine is 5.25 kg/s and these products exit from the engine at 560 m/s. If the pressure at the exhaust is approximately equal to the ambient air pressure, determine the thrust produced by the engine. 14.72: A rocket burns fuel at rate of 6.875 kg/s. The exhaust gases leave the rocket at a relative velocity of 1100 m/s and at atmospheric pressure. The gross weight of the rocket is 2200 N, the exhaust nozzle area is 325 cm2 and 2500 hp are developed by the rocket. Determine its velocity.

15 Dimensional Analysis and Similarity

Chapter Objectives •

To develop the rationale for dimensional analysis.

•

To introduce the Buckingham Pi theorem.

•

To apply the Buckingham Pi theorem to the case of rotating machines.

•

To establish criteria for geometric, kinematic, and dynamic similarity.

•

To employ similarity concepts in establishing scaling relationships for fluid mechanical systems.

15.1

Introduction

Dimensional homogeneity has been an important consideration in all of the material that has been presented thus far. Clearly, it is necessary that all quantitative relationships be dimensionally consistent and certain conversion factors are often used to achieve this consistency. In this chapter, we will use the principle of dimensional homogeneity in another way. By invoking a formalism to this concept, using dimensional analysis, we will combine important variables that describe a physical situation into a smaller number of dimensional groups. Such a process has a number of benefits. One benefit is that, in an experimental study involving a number of variables, the combination of variables into a smaller number of dimensionless parameters can greatly simplify the analysis and the presentation of experimental results. A second attribute of dimensionless groups is scaling. Scaling is the mathematical process of utilizing experimental results from a scale model to predict the performance of a fullsized device or system. A requirement of scaling is that the model and prototype be similar in all respects. The meaning and implications of scaling will be considered in detail in this chapter.

15.2

Fundamental Dimensions

In expressing the units of certain quantities, some have dimensions that are fundamental, and others are simply combinations of the fundamental dimensions. For example, if we consider mass, M, length, L, and time, T, as fundamental, we can express the dimensions 477

478

Introduction to Thermal and Fluid Engineering TABLE 15.1

Some Important Variables and Their Representations in the M, L , T System Variable

Symbol

Dimensions

Mass Length Time Area Volume

m L t A V Vˆ

M L T L2 L3

Velocity Acceleration Density Viscosity Force Torque Pressure Surface tension

a   F  P 

L/T L/T 2 M/L 3 M/L T ML/T 2 ML 2 /T 2 M/L T 2 M/T 2

pjwstk|402064|1435600742

of a large number of quantities as combinations of these. Table 15.1 presents a list of such quantities and their representation in the M, L , T system. We should note that the quantity, F, or force, appears on this list. Force and mass are related by Newton’s second law, F = ma , and conversion from force to mass involves the correction factor, gc . In some books dealing with the SI system, there is no reference to the factor, gc . Dimensional consistency in SI units is assured when proper units are employed for mass, force, displacement, and time. This is a consequence of the numerical value for gc being 1 (or unity). The reader is reminded that the conversion between mass and force does possess units even though the numerical value may be unity. In the SI system, gc has the numerical value of 1 kg-m/N-s2 and in the English system the numerical value is 32.17 lbm -ft/s2 -lbf . We could have made another choice and selected F, L , and T for the fundamental dimensions. In this case, mass would be represented in a list such as Table 15.1 with dimensions F T 2 /L. The choice of M, L , and T is arbitrary and we will use these as fundamental dimensions exclusively in this book. In a more extensive treatment of transport phenomena, we would need at least one more fundamental dimension: temperature, . In the fluid mechanics portion of this text (Chapters 11 through 18), we will be considering isothermal situations and, thus, M, L , and T will be sufficient for our purposes.

15.3

The Buckingham Pi Theorem

One means of generating dimensionless groups in a fluid mechanics application is to go through a formal procedure to nondimensionalize the governing differential equation for the process or phenomenon of interest. This approach will be discussed in Section 15.4. We often find that a system or device of interest is not easily described by a differential equation or, if it is, we don’t know its form ahead of time. An example of a complex system that is not amenable to description by differential equations would be a rotating machine such as a pump or turbine. The only possible way to evaluate such devices is by experimentation. Experimental results are then communicated in the form of dimensionless parameters. We will discuss rotating machinery in a more complete fashion in Chapter 18.

Dimensional Analysis and Similarity

479

For now we are interested in two questions regarding the appropriate dimensionless parameters: •

How many are there?

•

What are they?

Several methods exist for obtaining the answers to these questions and all of them have some common features. We will consider only one such procedure known as the Buckingham1 method. The method consists of two parts that relate, in turn, to answering the questions of “How many are there?” and “What are they?” The procedure for answering the question “How many are there?” involves the use of a straightforward rule known as the Buckingham Pi theorem, proposed in 1914. The designation, Pi, refers to the notation used for the dimensionless parameters, , which means “product of variables.” The idea is to first determine how many  groups are to be formed from the primitive variables that pertain to a given physical situation. The Buckingham Pi theorem is stated as i = n−r

(15.1)

where i = number of independent dimensionless  groups n = number of variables involved r = rank of the dimensional matrix The dimensional matrix is formed by tabulating the exponents on the fundamental dimensions, M, L , and T, which represent each variable involved. Example 15.1 provides an example of such a matrix and the evaluation of i, n, and r.

Example 15.1 Determine the number of dimensionless groups needed to represent the variables involved in describing the effects of flow normal to a cylinder. The variables ˆ the density, , and the viscosity, . involved are the force, F , the diameter, d, the velocity, V,

Solution The usual first step is to list all of the variables and their dimensional representation in the following form. Variable

Symbol

Dimensions

Force Diameter

F d Vˆ

ML/T 2 L

Velocity Density Viscosity

L/T M/L 3 M/L T

 

The second step is the formation of an array of exponents in the form

1 Edgar

F

d

Vˆ





M:

1

0

0

1

1

L:

1

1

1 −3

−1

T : −2

0

−1

0

−1

Buckingham (1867–1940) was an American physicist who worked in thermodynamics and dimensional analysis and who developed the Buckingham  theorem.

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Introduction to Thermal and Fluid Engineering

where the numbers coincide with the powers of M, L , and T in the dimensions of each variable. As an example, the dimensions of  are M/L T or M1 L −1 T −1 . Thus, the values of 1, −1, and −1 appear in the foregoing array. Doing this for all of the variables establishes the dimensional matrix which, for this problem, is ⎡ ⎤ 1 0 0 1 1 ⎢ ⎥ 1 −3 −1⎦ ⎣ 1 1 −2 0 −1 0 −1 This is a 3 × 5 matrix (one that contains 3 rows and 5 columns). The pi theorem is expressed in terms of the rank of this matrix. The rank of a matrix is the number of rows and columns in the largest nonzero determinant that can be formed from the matrix. In this case, the maximum value of r is three (r = 3), which can be verified by taking the first three columns of our matrix and obtaining the determinant ⎡ ⎤ 1 0 0 ⎢ ⎥ 1⎦ = −1 det ⎣ 1 1 −2

0 −1

which is nonzero. This shows that the rank, in this case, is 3. We note that r is also the number of fundamental dimensions. Some treatises express the pi theorem with r defined as the number of fundamental dimensions and, in a great majority of cases, this will be true. Exceptions, however, may occur and one will never experience difficulty if the pi theorem is interpreted as we have specified. Equation 15.1 now provides i = n−r = 5−3 = 2 which shows that there are two dimensionless groups. Our task is to determine the composition of the two dimensionless groups and Example 15.2 will demonstrate this process.

Example 15.2 From the results of Example 15.1, determine the combinations of variables that make up the two dimensionless parameters that are relevant to the case of flow normal to a cylinder.

Solution

We designate the  groups to be formed as 1 and 2 . Our procedure will be to first determine a core group of variables. The number of variables in the core will be r which, in our case, is 3. The process of evaluating the two  groups will consist of the following steps: 1. A core group of r variables (three for this example) will be chosen. 2. The core group will be combined with the remaining variables, one at a time. 3. The combination formed in step 2 will be made dimensionless by suitable choices of the exponents on the core variables. The variables in the core group must, among them, contain all of the fundamental dimensions. A feature of the process outlined in the foregoing list is that the core variables may appear in each dimensionless group. A guiding principle in choosing the core variables is to then exclude from the core those variables whose effect one wishes to isolate. In this

Dimensional Analysis and Similarity

481

example we are principally interested in the variation of the drag force, F , so we will not choose it to be in the core. The dynamic viscosity, , will be arbitrarily chosen as the other ˆ and  will then constitute the core variable to be excluded. The remaining variables, d, V, group. The two  groups will thus be 1 = d a Vˆ b c F and 2 = d a Vˆ b c  Our remaining task is to evaluate the exponents so that 1 and 2 are dimensionless. First, for 1 , we write 1 = d a Vˆ b c F Dimensionally, this becomes M0 L 0 T 0 = L a

 b  c L M ML 3 T L T2

We now equate the exponents on M, L , and T to form three simple linear algebraic equations in three unknowns M:0=

c +1

L : 0 = a +b −3c +1 T:0=

−b

−2

and the reader may verify that the solution of this set is a = −2,

b = −2,

and c = −1

These make 1 1 =

F d 2 Vˆ 2 

The second  group, 2 , can be formed in the same manner 2 = d a Vˆ b c  Dimensionally, this becomes  b  c L M M M L T =L 3 T L LT 0

0

0

a

and when we equate the exponents on M, L , and T, we obtain another set of three simple linear algebraic equations in three unknowns M:0=

c +1

L : 0 = a +b −3c −1 T:0=

−b

−1

The solution of this set yields the values a = −1,

b = −1,

and c = −1

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Introduction to Thermal and Fluid Engineering

These make our parameter 2 =

 ˆ d V

We see that, from dimensional analysis, in the case of a cylinder in cross flow, there are two applicable dimensionless parameters related functionally as 1 = f (2 ) Experimental data must be obtained for a determination of this function. We shall encounter this relationship in Chapter 16 where, the function, f will be determined. The two dimensionless groups that have been generated in this exercise are possibly new to the reader. In the fluid mechanics community, they, or similar forms, are often seen. The first of these groups, 1 =

F d 2 Vˆ 2 

can also be written in the form 1 =

P ˆ V2

where the numerator is the pressure, P = F /d 2 , with units of N/m2 . This parameter is considered, physically, to be the ratio of pressure forces to inertial forces and is given different symbols and names, depending on the branch of fluid mechanics involved. Often, it is designated as the Euler number symbolized as Eu. In future chapters, we will also see it designated as the drag coefficient, C D , and as the coefficient of skin friction, designated as C f . The second  group formed 2 =

 ˆ d V

2 =

ˆ Vd 

is most often seen in the form

and is designated as the Reynolds number symbolized as Re. The Reynolds number is, physically, the ratio of the inertial forces to viscous forces. The Reynolds number, named for Osborne Reynolds,2 a pioneer in the fluid mechanics field, is generally considered as the most important parameter in fluid mechanics. We will see these parameters, particularly the Reynolds number, again.

2 Osborne Reynolds (1842–1912) was a British engineer and physicist who demonstrated streamlines and turbulent

flow in pipes and showed the transition between them occurs at a critical velocity determined by the Reynolds number.

Dimensional Analysis and Similarity

15.4

483

Reduction of Differential Equations to a Dimensionless Form

An in-depth analysis of fluid mechanics generally involves the derivation and solution, where possible, of the governing differential equations. These equations also provide the basis for a considerable number of sophisticated computational codes. Such applications are beyond the scope of this text. When the differential equations that describes a given flow situation are known, they may be made nondimensional by using a relatively simple process that requires only a modest level of insight. A by-product of such a procedure is the formulation of dimensionless parameters associated with the flow. This process will now be illustrated. The starting equations for this exercise will not be derived. Their validity can be validated examining any of a large number of texts such as Welty et al. (2001) and Munson et al. (1998). A representative case to examine is two-dimensional compressible flow involving a velocity vector expressed in terms of its components V = Vˆ x x + Vˆ y y in a situation described by the equations ˆx ∂V ∂x

+

ˆy ∂V

(15.2)

∂y

and  

∂V

∂V ∂V + Vˆ x + Vˆ y ∂t ∂x ∂y



= g − ∇ P + 

∂2 V ∂x2

+

∂2 V ∂ y2

 (15.3)

Equation 15.2 is the continuity equation for two-dimensional incompressible flow. Its derivative is based on the principle of mass conservation, Equation 15.3 is a differential expression of Newton’s second law of motion. This equation is the two-dimensional form of the Navier3 -Stokes4 equations. Some physical insight is required at this point. We need to establish reference conditions for the flow and the following choices are made: •

Reference length = L

•

Reference velocity = Vˆ ∞

Nondimensional quantities (those with asterisks) in our differential equations will now be defined as follows: x∗ = Vˆ ∗x =

x L Vˆ x Vˆ ∞

y∗ = Vˆ ∗y =

y L Vˆ y Vˆ ∞

t∗ = Vˆ ∗ =

t Vˆ ∞ L Vˆ Vˆ ∞

∇ ∗ = L∇ 3 C.

L. M. H. Navier (1785–1836) was a French physicist and engineer who studied analytical mechanics and its application to the strength of materials, machines, and the motions of solid and liquid bodies. 4 G.

C. Stokes (1819–1903) was a British mathematician and physicist who originated the idea of determining the chemical composition of the sun and stars from their spectra and also studied double refraction and electromagnetic waves.

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Introduction to Thermal and Fluid Engineering

The last category in the foregoing tabulation, the gradient, is composed of first derivatives with respect to space coordinates. Hence, the length scale, L , is employed to make the gradient dimensionless. Our two differential equations may now be put into a dimensionless form. This process, in general, uses the chain rule of differentiation. For example, the continuity equation involves the terms  ˆx ˆ ∗ ∂Vˆ x ∂x ∗ ˆ∗ ∂V ∂V ∂V 1 Vˆ ∞ ∂Vˆ ∗x x x ˆ = = ( V ) = ∞ ˆ ∗ ∂x ∂x ∂x∗ ∂V ∂x∗ L L ∂x∗ x and ˆy ∂V ∂y

=

ˆ ∗ ∂Vˆ ∂ y∗ ∂V y y ˆ ∗ ∂y ∂ y∗ ∂ V y

=

Vˆ ∞ ∂Vˆ ∗y L ∂ y∗

which can be substituted into Equation 15.2 to give ˆ∗ ∂V x ∂x∗

+

ˆ∗ ∂V y ∂ y∗

=0

(15.4)

This dimensional equation has the same form in terms of nondimensional quantities as it had before. Utilizing the chain rule, as illustrated in the foregoing, the equation of motion is transformed to  ∗ ∗ Vˆ 2∞ ∂V ∗ ˆ x ∂V + Vˆ y ∂V + V L ∂t ∗ ∂x∗ ∂ y∗

 1 ∗ Vˆ ∞ ∂2 V ∗ ∂2 V ∗ = g − ∇ P + + ∗2 (15.5) L L2 ∂x ∗2 ∂y Note that every term in this equation has the units, M/L 2 T 2 or F /L 3 . Each term represents a certain kind of force, that is Vˆ 2∞ /L is an inertial force, Vˆ ∞ /L 2 is a viscous force, g is the gravitational force, and P/L is the pressure force. If we now divide by the quantity Vˆ 2∞ /L , our dimensionless equation becomes ∂V ∗ ∂t ∗

∂V ∗ ∂V ∗ + Vˆ x ∗ + Vˆ y ∗ ∂x ∂y

gL ∇∗ P  = − + 2 2 ˆ ˆ V∞ V ∞ L Vˆ ∞



∂2 V ∗ ∂x ∗2

+

∂2 V ∗ ∂ y∗2

 (15.6)

This equation looks very much like the original expression except that in the transformation to a dimensionless form, the gravitational and pressure terms have been modified and the coefficient of the viscous term is a group of variables. Moreover, an examination of these terms reveals that each is dimensionless. It is also possible to interpret these parameters, in physical terms, in light of the way they were developed. An initial consideration of the gravitational term gL/Vˆ 2∞ establishes, first, that it is indeed a dimensionless ratio. The choice of gL/Vˆ 2∞ or Vˆ 2∞ /gL is arbitrary—clearly both forms are dimensionless. The conventional choice is the latter; the Froude5 number (pronounced 5 William

Froude (1810–1879) was a British engineer who promulgated the Froude law of comparison when an object is towed in a liquid.

Dimensional Analysis and Similarity

485

Fr¯o o¯ d), defined by Fr ≡

Vˆ 2∞ gL

(15.7)

From its form, we may infer that the Froude number is a ratio of the inertial forces to the gravitational forces. The Froude number, an important scaling parameter in flows that involve a free surface, is encountered in the analysis of open channel flows. Another parameter associated with viscous terms, relating inertial forces and viscous forces, is the Reynolds number Re ≡

L Vˆ 2∞  

(15.8)

which we encountered in Example 15.2. A third dimensionless group is present in Equation 15.6. The pressure term now involves the group P/Vˆ 2∞ , which is dimensionless and can be interpreted as the ratio of the pressure forces to the inertial forces. Formally, this parameter is designated as the Euler6 number, which is given by Eu ≡

P Vˆ 2

(15.9)

∞

The Euler number was derived earlier using the Buckingham Pi method in Section 15.3.

15.5

Dimensional Analysis of Rotating Machines

One of the most important areas for applying the Buckingham Pi theorem is that of rotating machines such as compressors, fans, and turbines. These machines are sufficiently complex to preclude the possibility of a precise theoretical analysis. For a centrifugal pump, with the configuration shown in Figure 15.1, the important performance variables are listed in Table 15.2. We should note, before continuing, that some physical insight was necessary in developing the list of variables in Table 15.2. The variables  and  represent fluid properties that would be important in evaluating pump operation. The diameter, d, is a logical geometric parameter and the shaft speed (or angular velocity), n, which is normally controllable is ˙ are all performance also a straightforward choice. The other three variables, gh, P, and Q, quantities of interest. It is conventional practice in the pump industry to represent total head as gh in units of m2 /s2 rather than simply h in units of m. For a pump, operating on a system, the flow rate of fluid at a given pressure would naturally be of primary interest. It is important to know the power required to achieve this performance. Actual laboratory testing of a pump, with performance variables grouped according to dimensional analysis, is an important part of performance analysis that may lead to design modifications. 6 Leonhard Euler (1707–1783) was a Swiss mathematician who contributed to algebraic series and differential and

integral calculus and realized the significance of the Euler coefficients in certain trigonometric expansions.

486

Introduction to Thermal and Fluid Engineering

Casing Impeller Expanding Area Scroll FIGURE 15.1 Cutaway view of a centrifugal pump.

Now, to continue the dimensional grouping of the seven variables in Table 15.2, we first form the array gh

P

˙ V

d

n





M:

0

1

0

0

0

1

1

L:

2

2

3

1

0 −3

−1

−3 −1

0

T : −2

−1

−1

0

and this leads to the dimensional matrix ⎡

0

⎢ ⎢ 2 ⎣ −2

1

0

0

0

2

3

1

0

−3

−1

0 −1

1

1

⎤

⎥ −3 −1⎥ ⎦ 0 −1

The rank of this matrix is 3 and the pi theorem tells us that the number of dimensionless groups to be formed is i = n−r = 7−3 = 4 The next step is to choose the core group, which will consist of r = 3 variables. Clearly, ˙ to be in the core. One additional variable will be excluded from we do not want gh, P, or V the core and we choose the fluid viscosity, . With these choices, the four pi groups consist TABLE 15.2

Pump Performance Variables: Power Is ˙ Designated as P Rather Than W Variable Total head Power Flow rate Impeller diameter Shaft speed Fluid density Fluid viscosity

Symbol

Dimensions

gh P ˙ V d n  

L 2 /T 2 ML 2 /T 3 L 3 /T L 1/T M/L 3 M/L T

Dimensional Analysis and Similarity

487

of the following: 1 = d a nb c (gh) 2 = d a nb c P ˙ 3 = d a nb c V 4 = d a nb c  Here we evaluate 1 and ask the reader to verify the details for the evaluation of 2 through 4 . For 1 , we have 1 = d a nb c (gh) Dimensionally, this becomes M0 L 0 T 0 = L a (T −b )( Mc L −3c )(L 2 T −2 ) When we equate the exponents on M, L , and T, we obtain M: 0=c L : 0 = a − 3c + 2 T : 0 = −b + 2 The solution of this set yields the values a = −2,

b = −2,

and c = 0

These make our parameter 1 =

gh d 2 n2

pjwstk|402064|1435600783

Now, the reader may check the values of 2 through 4 . The resulting forms are 1 =

gh d 2 n2

2 =

P n3 d 5

3 =

˙ V nd 3

(head coefficient)

(power coefficient)

(flow coefficient)

(15.10)

(15.11)

(15.12)

and 4 =

 d 2 n

(a form of Reynolds number)

(15.13)

These four parameters are referred to as indicated. Typical pump performance curves are shown on coordinates with 1 and 2 plotted as a function of 3 . An example of such a performance plot is shown in Figure 15.2.

488

Introduction to Thermal and Fluid Engineering 0.25

100% η 80

0.20

60 CH

40

0.15

η

CH

20 0

0.10

0.016 CP (pump) 0.012 0.008

0.05

Cδ

0.004 0

0

0.025

0.050 CQ

0.075

0 0.100

FIGURE 15.2 Dimensionless pump performance curves.

15.6

Similarity

We have, on several occasions, alluded to the process whereby data acquired for a model of a device or system can be used to predict the performance of a full-scale operating entity. The idea of testing a model of an aircraft in a wind tunnel is obviously less expensive and faster than actually fabricating an aircraft and flight testing it. The question of scaling has to do with how we must interpret model behavior in order to predict the performance of the full-scale prototype. The concept of similarity is critical in this process. There are three types of similarity: •

Geometric similarity

•

Kinematic similarity

•

Dynamic similarity

Geometric similarity is an obvious requirement between a model and a prototype. A simple statement of geometric similarity involves the dimensions of the two systems being considered: Two systems are geometrically similar when the ratios of significant dimensions are equal for both. An example of the concept of two geometrically similar systems is illustrated in Figure 15.3. In Figure 15.3a, the two cylinders will be geometrically similar when L1 L2 = d1 d2

(15.14)

Dimensional Analysis and Similarity

489

L2 L1 d2

d1

(a) c2 c1

b1

b2

a1 a2 (b) FIGURE 15.3 Examples of geometrically similar systems (a) cylinders and (b) diamond shapes.

and, for the diamond-shape objects (Figure 15.3b), the requirement is a1 a2 = b1 b2

and

a1 a2 = c1 c2

or a1 b1 c1 = = a2 b2 c2

(15.15)

With regard to kinematic similarity: Two systems exhibit kinematic similarity when they are geometrically similar and when they are exposed to the same flows. Quantitatively, kinematic similarity corresponds to a case when, between systems 1 and 2, Vˆ y1 Vˆ x1 Vˆ z1 = = Vˆ x2 Vˆ y2 Vˆ z2

(15.16)

Here, it is presumed that the type of mean flow, either laminar or turbulent, applies to both systems. Finally, for dynamic similarity: Two systems exhibit dynamic similarity when they are geometrically and kinematically similar and when the ratios of significant forces between the systems are equal.

490

Introduction to Thermal and Fluid Engineering 100

Euler Number, P/V 2p

10

1

0.1

0.01 10–1

100

101

102

103

104

105

106

 Reynolds Number, pVd/μ FIGURE 15.4 Euler number variation as a function of Reynolds number for a cylinder in cross flow.

Because the dimensionless parameters generated earlier (such as the Reynolds and Euler numbers) physically represent such force ratios, dynamic similarity is exhibited when the parameters, which relate significant forces, are the same between the systems. In Example 15.2, it was determined that, in the case of flow across a cylinder, the dimensionless force ratios constituting the Reynolds and Euler numbers apply. Two such systems would thus be dynamically similar when

Re1 = Re2

and

Eu1 = Eu2

provided they are also geometrically and kinematically similar. This simple concept suggests that a plot of Eu as a function of Re, as shown in Figure 15.4, generated in a wellcontrolled experiment in a laboratory wind tunnel, can be used to predict the pressure drag force for any cylinder influenced by the flow of any fluid. The importance of such scaling is enormous. Example 15.3 will illustrate the use of scaling in predicting equipment performance.

Example 15.3 A common method for scaling cylindrical mixing tanks and impellers is to maintain a constant ratio of power to unit volume. If it is desired to increase the volume of a properly baffled liquid mixer by a factor of 5, by what ratio must the tank diameter and impeller speed be changed? We may presume that the tanks are geometrically similar and that both will operate in the fully turbulent regime. The power, P, supplied to the impeller may be assumed to be a function of the impeller diameter, d, its rotational velocity, n, and the liquid density, .

Solution The tank diameter ratio will be scaled by the requirement of geometric similarity. For cylindrical geometry, the two tanks, shown in Figure 15.5 have volumes

Va =

 2 d La 4 a

and

Vb =

 2 d Lb 4 b

Dimensional Analysis and Similarity

491 dB

dA

LB LA

(a)

(b)

FIGURE 15.5 Cylindrical mixing tanks for Example 15.3.

We may combine these expressions for the individual volumes in accordance with the stated relationship that Vb /Va = 5. Hence,  2 db L b Vb = 4 =5  2 Va da L a 4 or  2 db Lb =5 da La The condition of geometric similarity requires that da db = La Lb or Lb db = La da Making this substitution yields 

db da

3 =5

which gives us the required ratio of the diameters. db = 51/3 = 1.710 ⇐ da The ratio of impeller speeds is dictated by dynamic similarity requirements. Because we know the dependence of power on the characteristics of the impeller, P = f (d, n, ), we may combine these quantities into the dimensionless ratio P n3 d 5

492

Introduction to Thermal and Fluid Engineering

which was earlier designated as the power coefficient in Equation 15.11. For this parameter to be a constant between tanks a and b, we have Pa Pb = a na3 da5 b n3b db5 and for a constant power ratio, the speed ratio is nb = na



Pb Pa



da db

5

a b

1/3

= [(5)(5) −5/3 (1)]1/3 = 5−2/9 = 0.699 ⇐ The larger tank must have a diameter that is 1.71 times the diameter of the smaller one and an impeller speed that is approximately 0.70 times that of the smaller system. These results are not at all intuitive.

15.7

Summary

This chapter has introduced the formalism associated with the concepts of dimensional analysis, which is basically an application of dimensional homogeneity to a given situation. The Buckingham Pi theorem is a means to determine how many dimensionless groups may be formed from the appropriate listing of primitive variables that are significant in the analysis of a given flow situation. Once the number of  groups has been determined, a straightforward method for identifying the  groups may be implemented. The requirements of geometric, kinematic, and dynamic similarity enable one to use model data to predict the behavior of a prototype or full-size piece of equipment. Similitude or model theory is a very important application of the parameters obtained from dimensional analysis. Several applications of dimensional analysis will be considered in subsequent chapters.

15.8

Problems

Dimensionless Parameters and Dimensionless Forms 15.1: A car with a length of 5.8 m and a radio antenna of 6.4 mm is traveling along a road at 22.2 m/s. Determine the Reynolds number (a) based on the length of the car and (b) based on the diameter of the radio antenna. 15.2: Some common variables employed in the study of fluid mechanics include the vol˙ the acceleration of gravity, g, the viscosity, , the density, , umetric flow rate, V, and a length, L. Determine which of the following variables are dimensionless: (a) ˙ 2 /gL 2 , (b) V/L, ˙ ˙ and (d) VL/. ˙ V (c) gL 2 /V,

Dimensional Analysis and Similarity

493

15.3: In a gas, the velocity of sound, c, is known to depend on the pressure, P, and the density, . Use dimensional reasoning to establish (to within a constant) the relationship between c, P, and . 15.4: A cylindrical container is rotated around its vertical axix. The pressure, P, in the radial direction, depends on the rotational speed, , the location in the radial direction, r , and the density of the fluid, . Determine the form of the equation for P. 15.5: For the flow of a thin film of liquid with a depth, h, and a free surface, two imporˆ √gh and the Weber tant dimensionless parameters are the Froude number, Fr = V/ number, We = Vˆ 2 h/. Determine the value of both of these parameters for water at 20◦ C flowing with a velocity of 0.50 m/s at a depth of 3 mm. At 20◦ C, the surface tension of water,  = 0.0705 N/m. ˆ flows past an airfoil having a chord length, b, of 15.6: Standard air with a velocity, V, ˆ 2.5 m. Determine (a) the Reynolds number, Vb/ for Vˆ = 90 m/s and (b) if this airfoil were attached to an aircraft flying at the same speed in a standard atmosphere at an altitude of 3250 m. 15.7: The ratio of the viscous dissipation to the heat conduction in a fluid is given by the ˆ Brinkman number, Br. The Brinkman number is a combination of flow velocity, V, viscosity, , thermal conductivity, k and temperature, T. If the Brinkman number is proportional to viscosity, derive the Brinkman number. The thermal conductivity can be expressed dimensionally as k ∼ ML/T 3 , where  is temperature. 15.8: The velocity of propagation, C, of a capillary wave in deep water is a function of the wavelength, , the surface tension, , and the density, . Using a dimensionless constant, K , (a) determine the proper functional relationship for this phenomenon and (b) how the propagation velocity changes when the surface tension increases by a factor of 3. 15.9: Inside a soap bubble, the excess pressure, P, varies with surface tension, , and radius, r. Using dimensional analysis, determine how this pressure difference varies if (a) the radius increases by a factor of 1.5 and (b) the surface tension is doubled. ˙ of an ideal liquid through an orifice is a function of 15.10: The volumetric flow rate, V, the density of the liquid, , the diameter of the orifice, d, and the pressure difference across the orifice, P. Use dimensional analysis to determine the parameters involved in relating these variables. 15.11: Assuming the pressure is a function of density and velocity, determine the dynamic pressure exerted on an immersed object by a flowing incompressible fluid.

Reduction of Differential Equations to a Dimensionless Form 15.12: The flow between two concentric cylinders (Figure P15.12) is governed by the differential equation d 2 Vˆ  d + 2 dr dr



Vˆ  r

 =0

where Vˆ  is the tangential velocity at any local radial location, r. The inner cylinder is fixed and the outer cylinder rotates with an angular velocity, . Express the equation in dimensionless form using Ro and  as reference parameters.

494

Introduction to Thermal and Fluid Engineering ω

Ro Ri

FIGURE P15.12

15.13: An incompressible fluid is contained between two large parallel plates as shown in Figure P15.13. The upper plate is fixed, the fluid is initially at rest and if the bottom ˆ the differential equation plate suddenly starts to move with a constant velocity, V, governing the motion is 

ˆx ∂V ∂t

=

ˆx ∂2 V ∂ y2

where Vˆ x is the velocity in the x direction and  and  are the fluid density and viscosity, respectively. Rewrite the differential equation and the initial and boundary conditions in dimensionless form using h and Vˆ as reference parameters for length and velocity and h 2 / as a reference parameter for time. Fixed Plate

y

Vx

h

x  V FIGURE P15.13

15.14: The deflection of a cantilever beam shown in Figure P15.14 is governed by the differential equation EI

d2 y = P(x − L) d x2

where E is the modulus of elasticity and I is the moment of inertia of the beam cross section. The boundary conditions are y = 0 at x = 0 and dy/d x = 0 at x = 0. (a) Rewrite the equation and the boundary conditions in dimensionless form using the beam length, L , as the reference length and (b) based on the results of part (a), determine the equation to predict deflection.

Dimensional Analysis and Similarity

495

y

P x L FIGURE P15.14

15.15: A liquid is contained in a pipe that is closed at one end as shown in Figure P15.15. Initially, the liquid is at rest, but if the end is suddenly opened, the liquid starts to move. Assuming that the pressure, P1 , remains constant, the differential equation that describes the resulting motion of the liquid is

 ˆz P1 ∂2 V 1 ∂Vˆ z  = + + ∂t L ∂r 2 r ∂r ˆz ∂V

where Vˆ z is the velocity at any radial location, r , and t is time. Rewrite this equation in ˆ avg , as the reference dimensionless form using the pipe radius, R, the mean velocity, V ˆ dimensions. Note that the reference time becomes R/Vavg . End Initially Closed

P1 r

R z

L FIGURE P15.15

15.16: A plane wall of infinite extent in the y and z directions has the origin of its xcoordinate system at the centerline of the wall. The cooling of the wall of thickness 2L is governed by the Fourier equation ∂T ∂x2

=

1 ∂T

∂t

where is the thermal diffusivity of the wall material, T is the temperature, and t is the time. Express the differential equation in a nondimensional form using xo as the reference, T = Ti − T∞ as the reference temperature and xo / as the reference time. Use of the Buckingham π Theorem 15.17: Using the Buckingham Pi theorem, generate the dimensionless parameters that will correlate the variables in Problem 15.13. The velocity and time should be in separate parameters. 15.18: The power output of a hydraulic turbine depends on the diameter, d, of the turbine, the density, , of the water, the height, h, of the water surface above the turbine, the gravitational acceleration, g, the angular velocity, , of the turbine wheel,

496

Introduction to Thermal and Fluid Engineering ˙ of water through the turbine, and the efficiency, , of the turbine. the discharge, V, By dimensional analysis, generate a set of appropriate dimensional groups.

15.19: The pressure rise across a pump, P (this term is proportional to the head developed by the pump) may be considered to be affected by the fluid density, , the angular ˙ and the fluid velocity, , the impeller diameter, d, the volumetric rate of flow, V, ˙ viscosity, . Find the pertinent dimensionless groups, choosing them so that P, V, and  each appear in only one group. 15.20: The power, P, required to run a compressor varies with the angular velocity, , the ˙ the fluid density, , and the impeller diameter, d, the volumetric rate of flow, V, fluid viscosity, . Develop a relationship between these variables using dimensional analysis where fluid viscosity and angular velocity appear in only one dimensionless parameter. 15.21: The performance of a journal bearing around a rotating shaft is a function of the ˙ the lubricant following variables: the volumetric rate of flow of the lubricating oil, V, viscosity, , the fluid density, , the bearing diameter, d, the shaft speed in revolutions per minute, N, and the surface tension of the lubricating oil, . Suggest appropriate parameters that may be employed in the correlation of experimental data for such a system. 15.22: The mass, m, of drops formed by a liquid discharging by gravity from a vertical tube is a function of the liquid density, , the tube diameter, d, the surface tension, , and the acceleration of gravity, g. Neglecting any effects of viscosity, determine the independent dimensionless groups that would allow the surface tension effect to be analyzed. 15.23: The rate at which metallic ions are electroplated from a dilute electrolytic solution onto a rotating disk electrode is usually governed by the mass diffusion rate of ions to the disk. This process is believed to be controlled by the following variables: k = mass transfer coefficient, m/s D = diffusion coefficient, m2 /s d = disk diameter, m  = angular velocity, 1/s  = density, kg/m3  = viscosity, kg/m-s Obtain a set of dimensionless groups for these variables keeping k, , and D in separate groups. 15.24: The fundamental frequency, n, of a stretched string is a function of the string length, L , its diameter, d, the mass density, , and the applied tensile force, T. Suggest a set of dimensionless parameters relating these variables. 15.25: A thin elastic wire is placed between supports. A fluid flows past the wire and it is desired to study the static deflection, , at the center of the wire due to the fluid drag. Assume that ˆ E) = f (L , d, , , V, where L is the length of the wire, d is the wire diameter,  is the fluid density,  is the fluid viscosity, Vˆ is the fluid velocity, and E is the modulus of elasticity of the wire material. Develop a suitable set of pi terms for this problem.

Dimensional Analysis and Similarity

497

15.26: Through a series of tests on pipe flow, Darcy7 derived an equation for the friction loss in pipe flow as hL = f

L Vˆ 2 d 2g

in which f is a dimensionless coefficient that depends on the average velocity of ˆ the pipe diameter, d, the pipe length, L, the fluid density, , the the pipe flow, V, fluid viscosity, , and the average pipe wall roughness or unevenness, e, which is a length term. Using the Buckingham Pi theorem, find a dimensionless function for the coefficient, f. 15.27: The maximum pitching moment that is developed by the water on a flying boat as it lands is denoted as Cmax . The following are the variables involved in this action:

= angle made by the flight path of the aircraft with the horizontal

= angle defining the attitude of the aircraft m = mass of the aircraft L = length of the hull  = density of water g = acceleration of gravity R = radius of gyration of the aircraft about the pitching axis Use the Buckingham Pi theorem to determine (a) the number of independent dimensionless groups needed to characterize this problem, (b) the dimensional matrix of the problem, and (c) the rank of the dimensional matrix. ˙ of a gas from a smokestack is a function 15.28: Assume that the volumetric flow rate, V, of the density of ambient air, a , the density of the gas within the stack, g , the acceleration of gravity, g, the height of the stack, h, and the diameter of the stack, d. Use a , d, and g as repeating variables to develop a set of pi terms that could be employed to describe this problem. 15.29: In a fuel injection system, small droplets are formed due to the breakup of the liquid jet. Assume that the droplet diameter, d, is a function of the liquid density, , the fluid viscosity, , the surface tension, , the jet velocity, Vˆ and the diameter, d. ˆ and d as repeating Form an appropriate set of dimensionless parameters using , V, variables. ˆ through a hole in the side of a large tank. Assume 15.30: A liquid flows with a velocity, V, that Vˆ = f (h, g, , ) where h is the depth of the fluid above the hole, g is the acceleration of gravity,  is the fluid density and  is the surface tension. The following data were obtained by changing h and measuring Vˆ with a fluid having a density of 1000 kg/m3 and a surface tension of 0.74 N/m: ˆ m/s V,

3.13 4.43

5.42

6.25

7.00

h, m

0.50 1.00

1.50

2.00

2.50

7 H. P. G. Darcy (1803–1858) performed extensive tests on filtration and pipe resistance and initiated open-channel

studies.

498

Introduction to Thermal and Fluid Engineering Plot these data by using appropriate dimensionless variables and comment as to whether any of the original variables could have been omitted.

15.31: As indicated in Figure P15.31, the pressure drop across a short hollowed plug placed in a circular tube through which a liquid flows can be expressed as ˆ D, d) P = f (, V, where  is the fluid density and Vˆ is the mean velocity in the tube. Some experimental data obtained with D = 61 cm,  = 1031 kg/m3 , and Vˆ = 0.61 m/s are given in the following table: d, cm P, N/m

2

1.829

2.438

3.048

4.572

23, 643

7479

3064

603

Plot these data by using appropriate dimensionless variables. ΔP

V

D

d

FIGURE P15.31

15.32: One type of viscosimeter consists of an open reservoir with a small diameter tube at the bottom as illustrated in Figure P15.32. To measure the viscosity, the system is filled with the liquid of interest and the time required for the liquid to fall from h i to h f is determined. Variables involved are the initial head, h i , and the final head, h f (h = h i − h f ), the density, , the viscosity, , the time, t, the acceleration of gravity, g, and the diameter, d. Determine the dimensionless parameters that would be appropriate in presenting test measurements for this system.

hi

d

hf

FIGURE P15.32

15.33: The concentric cylinder device of the type shown in Figure P15.33 is commonly used to measure the viscosity, , of liquids by relating the angle of twist, , to the angular

Dimensional Analysis and Similarity

499

velocity, , of the outer cylinder. Assume that  = f (, , K , d1 , d2 , L) where K depends on the suspending wire properties and has the dimensions, F L . The following data were obtained in a series of tests for which  = 0.01488 kg/m-s, K = 13.588 J, L = 0.3048 m, and d1 and d2 were constant: , rad

, rad/s

0.89

0.30

1.50

0.50

2.51

0.82

3.05

1.05

4.28

1.43

5.52

1.86

6.40

2.14

From these data, determine, with the aid of dimensional analysis, the relationship between , , and  for this particular apparatus. Hint: Plot the data, using appropriate dimensionless parameters, and determine the equation of the result using a standard curve-fitting technique. The equation so obtained should satisfy the condition that  = 0 for  = 0. Fixed Support Q

Rotating Outer Cylinder

Liquid

Inner Cylinder

. L

d1 d2 FIGURE P15.33

15.34: An estimate is needed of the lift provided by a hydrofoil wing section when it moves through water at 27.5 m/s. Test data are available for this purpose from experiments in a pressurized wind tunnel with an airfoil section model geometrically similar to but twice the size of the hydrofoil. The lift, F , is a function of the density of the ˆ the angle of attack, , the chord length, d, and fluid, , the velocity of the flow, V, the viscosity, . Take the density of water as 1000 kg/m3 and assume that the angle of attack is the same in both cases, the density of the air in the pressurized wind tunnel is 2.58 kg/m3 , the kinematic viscosity of the air is 7.43 × 10−6 m/s2 , and the kinematic viscosity of the water is 0.929 × 10−6 m/s2 . Determine the velocity of flow in the wind tunnel that would correspond to the hydrofoil velocity for which the estimate is desired.

500

Introduction to Thermal and Fluid Engineering

Similarity 15.35: The pressure drop in a Venturi meter is known to vary only with the fluid density, ˆ and the diameter ratio of the meter. Meter 1 in water  the velocity of approach, V, shows a 6.25-kPa drop when the velocity of approach is 4.875 m/s. Meter 2, which is geometrically similar to meter 1 is used with benzene ( = 682 kg/m3 ) in an ˙ = 0.1688 m3 /s. Determine the pipe diameter of meter 2 that application where V will yield a 16.25-kPa pressure drop in meter 2. 15.36: In order to find the drag on a 0.10-cm-diameter sphere in slowly flowing water at 20◦ C a 0.925-cm-diameter sphere is tested in glycerin at 320 K and Vˆ = 0.2875 m/s with a measured drag on the model of 1.325 N. Determine the water velocity and the drag on the smaller sphere in water. 15.37: It is proposed that a subsonic wind tunnel be used to determine the drag on an aircraft whose velocity is 125 m/s using a 1:24 scale model at the same pressure and temperature. Can this be done? 15.38: A ship that is 195-m long is to operate at 10 m/s in seawater ( = 12.61 × 10−6 m/s2 ). Determine the kinematic viscosity of a fluid employed with a 3.35-m model that preserves the equality of both the Reynolds and the Froude numbers. 15.39: A model of a ship propeller is to be tested in water at the same temperature that would be encountered by a full-scale propeller. Over the speed range considered, it is assumed that there is no dependence on either the Reynolds or the Euler numbers but ˆ and propeller diameter, only on the Froude number (based on the forward velocity, V, ˆ d). In addition, the ratio of the forward to rotational velocity of the propeller (V/nd) where n is the propeller rpm, must be constant. With a model 0.41 m in diameter, a forward speed of 2.58 m/s and a rotational speed of 450 rpm, determine the forward and rotational speeds of a 2.45-m prototype. 15.40: A 40% scale model of an aircraft is to be tested in a flow regime where unsteady flow effects are important. The full-scale vehicle experiences the unsteady effects at a Mach number of 1.00 at an altitude of 12,000 m. The model is to be tested at 22◦ C and noting that the speed of sound is proportional to the square root of the temperature, determine the temperature that the model must be tested at to produce an equal Reynolds number. 15.41: During the development of a ship having a length of 125 m, it is desired to test a 10% model in a towing tank to determine the drag characteristics of the hull. Determine how the model is to be tested if the Froude number is to be duplicated. 15.42: A 25% scale model of an undersea vehicle that has a maximum speed of 16 m/s is to be tested in a wind tunnel with a pressure of 6 atmospheres to determine the drag characteristics of the full-scale vehicle. The model is 3-m long. Determine (a) the air speed required to test the model and (b) the ratio of the model drag to the fullscale drag. 15.43: A one-sixth scale model of a torpedo is tested in a water tunnel to determine drag characteristics. Determine (a) the prototype resistance if the model resistance is 45 N and (b) the velocity that corresponds to a torpedo velocity of 20 knots. 15.44: For a certain fluid flow problem, it is known that both the Froude and Weber numbers are important dimensionless parameters. The problem is to be studied using a 1/10 scale model and the model and the prototype operate in the same gravitational field. Determine the required surface tension scale if the density scale is equal to 1.00. The Weber number, We, relates inertial forces to surface tension forces and has the form We = Vˆ 2 L/.

Dimensional Analysis and Similarity

501

15.45: A large, rigid, rectangular billboard is supported by an elastic column as shown in Figure P15.45. There is concern about the deflection, , of the top of the structure durˆ A wind-tunnel test is to be conducted with a 1/15 scale ing a high wind of velocity, V. model. Assume that the pertinent column variables are its length, cross-sectional dimensions, and the modulus of elasticity of the material used for the column. The only important “wind” variables are the air density and the velocity. Determine (a) the model design conditions and the prediction equation for the deflection and (b) if the same structural materials are used for both the model and the prototype and the wind tunnel operates under standard atmospheric conditions, find the required wind-tunnel velocity to match an 80-km/h wind. V Billboard

Front View

Side View FIGURE P15.45

15.46: The pressure drop between the entrance and exit of a 150-mm-diameter 90◦ elbow through which ethyl alcohol at 300 K is flowing is to be determined with a geometrically similar model. The velocity of the alcohol is 5 m/s and the model fluid is to be water at 20◦ C with a velocity limited to 10 m/s. Determine (a) the required diameter of the model elbow to maintain dynamic similarity and (b) the prototype pressure drop if the measured pressure drop in the model is 20 kPa. 15.47: A solid sphere having a diameter, d, and a density, s , is immersed in a liquid having a density,  f ( f > s ) and then released. It is desired to use a model system to determine the maximum height, h, above the liquid surface to which the sphere will rise, upon release, from a depth, H. It can be assumed that the important liquid properties are the density,  f and the viscosity,  f . Establish the model design conditions and the prediction equation for the system. 15.48: The drag characteristics for a newly designed automobile having a maximum characteristic length of 6.10 m are to be determined through a model study. The characteristics at both low speed (approximately 8.94 m/s) and high speed (44.70 m/s) are of interest. For a series of projected model tests, an unpressurized wind tunnel will be used to accommodate a model with a maximum characteristic length of 1.21 m. Determine the range of air velocities that would be required for the wind tunnel if Reynolds number similarity is desired and comment on the suitability of the velocities. 15.49: Determine the drag due to an 80-km/h wind on a 2-m-diameter satellite dish. A wind-tunnel test is to be performed using a geometrically similar 0.5-m-diameter model dish. Assume standard air for both model and prototype and determine (a) the air speed at which the model should be run and (b) the predicted drag on the prototype dish with all similarity conditions satisfied and a measured drag on the model of 200 N. What is the predicted drag on the satellite dish?

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15.50: An 8-cm sphere is tested in 30◦ C water flowing at 5 m/s, yields a measured drag of 12 N. For a 2.875-m weather balloon moving in air at 7◦ C and one atmosphere, determine the velocity and drag force. 15.51: Is it reasonable to use air in a subsonic wind tunnel at the same temperature and pressure for a test to find the drag on a 1:24 scale model of an aircraft whose velocity is 4.5 km/h. 15.52: A 1:24 scale model of a submarine is tested at 60 m/s in a wind tunnel using standard air at sea level. (a) Determine the prototype speed in seawater at 20◦ C and (b) for a model drag is 7.20 N, determine the prototype drag. 15.53: A model airplane represents a 1:24 scale prototype. The prototype is required to fly at 765 km/h, the Reynolds numbers of both model and prototype are required to be identical and the air temperature and pressure are to remain the same. Determine the air velocity in the wind tunnel. 15.54: A freighter that is 200-m long must operate at 0.975 m/s in seawater whose kinematic viscosity is 0.9982×10−6 m/s2 . Determine the kinematic viscosity of the fluid used in a test of a 3.50-m model so that both the Reynolds and Froude number are equivalent. 15.55: To assess overall head losses, a model of a venturi meter has linear dimensions that are 15% of those of the prototype. Both the model and the prototype operate in water, the model at 10◦ C and the prototype at 100◦ C. For a throat velocity of 6.75 m/s and a throat diameter of 67.5 cm, determine the flow through the meter. 15.56: Air at 20◦ C at a velocity of 2.15 m/s flows through a 6-cm pipe. For dynamic similarity, determine the size of pipe to carry water at 15◦ C and a velocity of 1.15 m/s. 15.57: A blimp is designed to move in air at 27◦ C at 6.75 m/s. A 1:20 scale model is used in water at 20◦ C with a measured drag of 3.00 kN. Determine (a) the water velocity of the model and (b) the drag on the prototype, and (c) the power required for its propulsion. 15.58: A 1:8 scale model of an automobile is tested in a wind tunnel having the same properties as the prototype. The velocity of the prototype is 45 km/h and, for dynamically similar airflow conditions, the model drag is 350 N. Determine the drag on the prototype and the power required for model propulsion. 15.59: A one-sixth scale model of a torpedo is tested in a wind tunnel. The prototype is expected to attain a velocity of 6.25 m/s in water at 60◦ C. Determine the model speed for the wind-tunnel pressure of 2 MPa and its temperature 80◦ C. 15.60: Water at 20◦ C flows at 4 m/s in a 15-cm-diameter pipe. For dynamic similarity, determine the velocity of a fuel oil having a kinematic viscosity of 8.012 × 10−6 m/s2 in a 7.5-cm-diameter pipe. 15.61: Air at 27◦ C is to flow through a 60-cm-diameter pipe at an average velocity of 2.25 m/s for dynamic similarity to be maintained. Determine the size of pipe required to carry 20◦ C water at 1.25 m/s.

16 Viscous Flow

Chapter Objectives •

To review Reynolds’ classic experiment.

•

To consider the boundary layer concept and it implications.

•

To develop concepts relative to viscous flow over bluff bodies.

•

To establish concepts relative to viscous flow along plane surfaces.

16.1

Introduction

In this chapter, we will combine some of ideas that have been developed earlier to describe the total interaction between a solid body and a fluid flowing through it or around it. The chapter title, “Viscous Flow,” correctly implies that we will be considering viscous effects in contrast to the analysis of inviscid flow, which is a subject of particular interest to aerospace engineers. The subject of inviscid flow tends to be quite sophisticated mathematically and is beyond the scope of this book. One of the major success stories in the annals of fluid mechanics is the concept of the boundary layer proposed by Ludwig Prandtl1 around the beginning of the twentieth century. The idea of a thin region, near a solid boundary, where viscous flow effects are manifested, has led to much of the useful modeling on which we now rely to describe fluid flow effects. Outside the boundary layer, viscous flow effects are negligibly small and the flow can be idealized to follow the Bernoulli equation. We will now discuss some practical aspects of viscous flow.

16.2

Reynolds’ Experiment

The existence of two kinds of viscous flows, laminar and turbulent, is a universally accepted concept. Water flowing slowly from a faucet proceeds smoothly and uniformly for a short distance and, then, abruptly changes into an unstable, irregular behavior. Similar behavior can be observed for smoke emanating from a slowly burning object.

1 L.

Prandtl (1875–1953) was a German physicist who contributed to fluid mechanics, particularly aerodynamics, and introduced the concept of the boundary layer.

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The well-ordered type of flow has already been designated as laminar to indicate the layer-like or laminar behavior as adjacent fluid layers slide smoothly over one another without significant mixing between them. The other flow behavior, in which packets of fluid particles move between layers giving it a fluctuating character, is designated as turbulence and the result is turbulent flow. In 1883, Osborne Reynolds first described an experiment that, quantitatively, established the transition between laminar and turbulent flow. His experiment, which has been duplicated countless times, was conducted in an apparatus with general features as shown in Figure 16.1a. Water was allowed to flow through a transparent tube at rate controlled by the valve. A dye with a specific gravity of unity, to avoid buoyancy effects, was introduced into the flow at a location near the flow entrance, and the appearance of the dye streak was observed as the valve opening was increased to provide greater flow rates. As the valve was opened slowly, the dye streak retained its regular flow character, forming a single line of color as illustrated in Figure 16.1b (top). As the valve opening was increased, producing ever larger flow rates, it was observed that the dye suddenly became dispersed through the cross section (Figure 16.1b bottom). This behavior was clearly due to the orderly character of laminar flow in the first case and the chaotic nature of turbulent flow in the latter case. It was clear, in Reynolds’ experiment, that the transition from laminar to turbulent flow was velocity dependent. In investigating other variables affecting this transition, he found that pipe diameter, fluid density, and fluid viscosity were also important. The combination of these four variables into a dimensionless parameter produced the Reynolds number Re =

ˆ Vd 

(16.1)

named in honor of the British engineer, Sir Osborne Reynolds, whose experiment first demonstrated its importance. The Reynolds number is often specified on the basis of the length dimension employed. For example, when the Reynolds number is based on the diameter, Red is often used; when a length or a length coordinate is employed, we frequently see the Reynolds number indicated by Re L or Rex . The transition from laminar to turbulent flow in a pipe is generally agreed to occur at a value of Re equal to 2300. Below this value, the flow is definitely laminar. For carefully controlled experiments, laminar flow has been reported for values of Re as high as 40,000. However, at high values of Re, a slight disturbance will cause the flow to become turbulent. Laminar flows at a Reynolds number above 2300 are therefore unstable and this value, 2300, is the generally accepted value for the critical Reynolds number for pipe flow.

Dye

Pipe

D

Laminar

Dye Streak

Turbulent

Smooth, Well-rounded Entrance (a) FIGURE 16.1 (a) Features of the Reynolds’ experiment and (b) flow behaviors.

(b)

Viscous Flow

16.3

505

Fluid Drag

In Example 15.1, in the preceding chapter, we used dimensional analysis to determine that the effects of flow normal to a cylinder are related by two parameters 1 =

F /d 2 Vˆ 2

and

2 =

 ˆ d V

The second of these dimensionless quantities, 2 , is observed to be 1/Re or we could simply use Re, which is the better-accepted dimensionless form. The other dimensionless parameter, 1 , relates the resisting or drag force per length squared (we might think of this as an area) to the kinetic energy of flow. An equivalent dimensionless form is F Vˆ 2 = 1 ∞ A 2 where F = the drag force A = the area  = the fluid density Vˆ ∞ = the free-stream velocity Note that the kinetic energy has been written with a 2 in the denominator, which is the generally accepted form. The coefficient, 2 , which can be thought of as a proportionality factor, is generally given the symbol, C. These coefficients are commonly encountered in two forms. One, designated C f , is the coefficient of skin friction. It relates the viscous or frictional drag force to the kinetic energy of flow according to the expression F Vˆ 2∞ = Cf A 2

(16.2)

where F = the frictional drag force A = the surface area of the solid body contacting the flowing fluid C f = the coefficient of skin friction As we will see directly, Equation 16.2 applies for flow over streamlined bodies where flow generally conforms to the shape of the solid boundary. The other coefficient, designated as C D , is termed the drag coefficient and relates the drag force due to pressure differences fore-and-aft to the kinetic energy of the flow. The relationship that applies in this case is F Vˆ 2 = CD ∞ A 2 where F = the total drag force A = the projected area of the body normal to the direction of flow C D = the drag coefficient

(16.3)

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Introduction to Thermal and Fluid Engineering 100

CD

10

1

0.1 10–1 2

4 6 8 100

2

4 6 8 101

2

4 6 8 102

2

4 6 8 103

2

4 6 8 104

2

4 6 8 105

2

4 6 8 106

 Re = Vd/u FIGURE 16.2 Drag coefficient variation with the Reynolds number for flow across a circular cylinder.

The drag force given by Equation 16.3 is generally associated with bluff (nonstreamlined) bodies where the streamlines actually separate from the surface giving rise to significant pressure differences and, consequently, significant drag forces. It should also be noted that the measured drag force for flow over bluff bodies includes both those due to friction and pressure. The magnitude of viscous drag is usually much the smaller of the two. The drag coefficient, as defined by Equation 16.3, may be thought of as that due to pressure effects but does, in fact, include the relatively smaller viscous effects as well. Recall that dimensional analysis indicates these coefficients are both functions of the Reynolds number. Figure 16.2 shows the variation in C D with Red for flow normal to a circular cylinder. The characteristics shown are the results of many experimental studies. It is important that we understand the mechanical phenomena responsible for the behavior illustrated in Figure 16.2. We will discuss this behavior, proceeding as the flow velocity (thus Red ) increases through four regimes.

2 T.

•

Regime I: At very small Reynolds numbers, generally in the regime Red < 1, the condition is often designated as one of creeping flow. Here, the flow is completely laminar and is slow enough to completely follow the shape of the body. Under these conditions, drag is totally due to viscous effects that extend throughout the flow field. Flow is stable and the downstream flow known as the wake does not oscillate. Typical flow patterns for a cylinder in cross flow are shown in Figure 16.3. The creeping flow case is represented in regime I.

•

Regime II: In Regime II, which extends in the Red range from 1 to approximately 103 , small eddies begin to form near the aft stagnation point. As the velocity increases, the eddies also grow in extent and are shed, alternately, from the upper and lower portions of the separated region. As they are swept downstream, they impart an oscillating character to the wake. At values of Red where this oscillatory behavior is distinct, the pattern is referred to as a Von Karman2 vortex “street”

Von Karman (1881–1963) was an American aerodynamicist who made theoretical contributions to aerodynamics and who formulated von Karman’s theory of vortex sheets, an early step in the mathematical treatment of turbulent motion.

Viscous Flow

507 100

CD

10

1

0.1 Regimes I 0.01 10–1

100

II 101

III 102

103

104

IV 105

106

 Reynolds Number = pVd/μ FIGURE 16.3 Drag coefficient variation with the Reynolds number for flow across a circular cylinder.

or trail. In this region, the wake is unsteady and the streamlines actually separate from the rear of the cylinder; , the angle between the forward stagnation point and the separation point, becomes smaller for increasing values of Red . In this regime, the drag coefficient decreases in a regular manner as Red increases. •

Regime III: As the velocity increases, the separation point continues to move forward, stabilizing at a position approximately 80◦ from the forward stagnation point. In this region, extending roughly between Red values from 103 to 3 × 105 , the value of C D remains near a value of 1. Over the portion of the surface where the streamlines are attached, the flow is laminar. The wake region is no longer characterized by large eddies but remains stable.

•

Regime IV: An interesting and important phenomenon occurs for a value of Red near 5 × 105 , where the drag coefficient suddenly decreases from a value near 1 to approximately 0.3. The explanation for this behavior is that boundary layer flow in the attached region becomes turbulent and, in turn, the separation point moves rearward from  = 80◦ to approximately 145◦ . As a result of this observation, we may conclude that a turbulent boundary layer tends to remain attached longer or, in other words, resists separation more, than does a laminar boundary layer. The direct result of a greater region of boundary layer attachment is the sudden decrease in C D that is evident in Figure 16.2. As Red increases beyond the transition value, C D will, again, increase slowly.

Figure 16.4 illustrates the pressure distribution about the cylindrical surface for values of Red typical of the flow regimes discussed. For the case of attached streamlines, the pressure distribution over the front and back portions of the cylinder is seen to be the same. Thus, there is no net pressure force. However, when boundary layer separation occurs, the effective pressure force over the forward portion of the cylinder is greater than over the rear and a net pressure force will exist. There will still be some viscous drag contributed by the region where the boundary

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Introduction to Thermal and Fluid Engineering 2

θ

V∞ P∞

1

2(P – P∞) PV 2o

0 Re = 105 –1

–2

Re = 6 × 105

–3 Inviscid Flow

–4

0

30

60

90 120 θ, Degrees

150

180

FIGURE 16.4 Pressure distribution over a cylinder in cross flow for different values of Red .

layer is attached but this will be small compared with the drag due to the pressure imbalance. When separation exists the coefficient used to evaluate drag is, thus, C D . The pressure distributions for laminar and turbulent boundary layer flows show why the value of C D for the turbulent case is lower than for the laminar case. This is a direct consequence of boundary layer separation. The situation for flow past a sphere is very much like that just discussed for the case of the cylinder except that, at the same values of Red , C D will differ between the two shapes. Figure 16.5 shows C D as a function of Red for three other plane surface cases oriented perpendicular to the free-stream velocity. We note that for Red < 1, the asymptotic limit for C D in the case of spheres is described by the expression

CD =

24 Red

This is the Stokes’ limit or Stokes’ law, achieved analytically by Stokes. Table 16.1 gives values of C D for a number of bluff objects.

Example 16.1 Evaluate the steady-state terminal velocity of a glass sphere measuring 1 cm in diameter falling freely through (a) glycerin at 300 K, (b) water at 300 K, and (c) air at 300 K. The density of glass may be taken as 2250 kg/m3 .

Viscous Flow

509

200 100 50 20

CD

10 Stokes’ Drag 5 CD = 24/Re

Sphere

V∞

d Spheres

V∞

d Infinite plates

V∞

d Circular disk and square plates

Infinite Plates

2 1 0.5

Circular Disks and Square Plates

0.2 0.1 0.05 0.02 0.1 0.2 0.5 1 2

5 10 20 50 100 200 500 1000

10,000

100,000

1,000,000

 Reynolds Number = pVd/μ FIGURE 16.5 Drag coefficients as a function of Red for various bluff bodies.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The glycerin is incompressible. A freely falling object will reach a steady, terminal velocity when the forces due to its weight and to fluid drag are in balance. The expression to be used equates the weight s

d 3 g 6

with the drag C D f

Vˆ 2∞ d 2 2 4

TABLE 16.1

Drag Coefficient Values for Various Blunt Shapes: The Value for the Ring is Based on the Ring Area Shape

CD

Disk Ring Finite Cylinder w/ h = 1 w/ h = 10 w/ h = 20 Hemisphere Open end facing flow Open end facing downstream Half-Cylinder Open end facing flow Open end facing downstream

1.17 1.20 1.12 1.31 1.44 1.42 0.38 2.30 1.20

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Introduction to Thermal and Fluid Engineering

to yield 4 s C D Vˆ 2∞ = dg 3 f We may recall from Figure 16.5, that the drag coefficient for a sphere is a function of the Reynolds number, Red . Because Red is a function of Vˆ ∞ , it is not possible to solve this expression explicitly for Vˆ ∞ unless Red < 1, which would permit the use of Stokes’ law in expressing C D . In this case, we must conclude that a trial-and-error solution is required. The conditions to be satisfied are the force balance and the figure that gives C D as a function of Red . The results for the three fluids are (a) For glycerin C D ∼ =5 Red ∼ = 8.6 Vˆ ∞ ∼ = 0.305 m/s (b) For water C D ∼ = 0.42 Red ∼ = 3.1 × 104 Vˆ ∞ ∼ = 1.34 m/s (c) For air C D ∼ = 0.46 Red ∼ = 4.19 × 106 Vˆ ∞ ∼ = 32.9 m/s An interesting example of the application of these ideas is the golf ball, with its dimpled surface. The purpose of this deliberate roughing of the surface is to induce turbulence in the boundary layer so that boundary layer separation will be reduced and the drag force will, in turn, be decreased. Because its drag is reduced, a dimpled golf ball will travel significantly further than would a smooth one. Design Example 8, which now follows, shows how the power required to overcome vehicle drag may be determined.

16.4

Design Example 8

The drag coefficient for a late model automobile having a frontal area of 2.354 m2 is specified as C D = 0.28. (a) Plot the power necessary to overcome aerodynamic drag as a function of vehicle speed from 40 mph to 80 mph in the air at 16◦ C and (b) determine the vehicle speed for an applied 20 hp.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The surrounding air is incompressible.

Viscous Flow

511 TABLE 16.3

Summary of Computations to Determine the Power Needed to Overcome Vehicle Drag Assume Vˆ Vˆ Vˆ 3 ˙ W ˙ W

mph

40

50

60

70

80

m/s

17.88

22.35

26.82

31.29

35.76

m3 /s3 kW hp

5717 2.32 3.11

11,166 4.52 6.06

19,292 7.81 10.48

30,635 12.41 16.64

45,729 18.52 24.84

(a) For air at 16◦ C (289 K), Table A.13 in Appendix A gives (with interpolation)  = 1.229 kg/m3 The drag force on a bluff body is related to the velocity according to Fd = AC D 

Vˆ 2 2

and the power required is  ˙ = Fd Vˆ = W

Vˆ 2 AC D  2



Vˆ 3 Vˆ = AC D  2

We observe that the power varies as the third power of the velocity. ˙ in Table 16.3 provides a summary of the data needed to be generated in order to plot W hp as a function of velocity. These data are plotted in Figure 16.6. (b) For a 20-hp application, the speed of the vehicle is approximately 74 mph.

Power to Overcome Drag, hp

30

20

10

40

50

60 70 Velocity, mph

80

FIGURE 16.6 Power required to overcome drag as a function of vehicle velocity for a late-model automobile.

512

16.5

Introduction to Thermal and Fluid Engineering

Boundary Layer Flow over a Flat Plate

A classic problem in fluid mechanics is the case of flow associated with a plane surface oriented parallel to the direction of flow. The boundary layer concept of Prandtl, dating to 1904, is that all viscous effects are confined to a thin region close to the solid surface. In this boundary layer region, the velocity changes from zero at the wall (owing to the no-slip condition) to the free-stream value at its outer edge. Figure 16.7, illustrates the velocity profiles at several positions along a flat plate. This figure is not to scale. In reality, the boundary layer region is much thinner than as shown in the figure where the vertical scale has been exaggerated for clarification. In examining these velocity profiles, certain qualitative conclusions may be reached:

(a) The boundary layer region grows in a regular manner with increasing distance, x, from the leading edge. (b) The velocity profiles near the leading edge are similar in shape, the major difference between them being the distance from the surface where the velocity reaches the ˆ ∞ . The boundary layer flow in this region is laminar. free-stream value, V (c) At large values of x, there is a distinct change in the shape of the velocity profiles. Near the wall, the velocity changes rapidly; then the slope of the profile becomes ˆ ∞ . This is the turbulent quite small for a considerable distance until, finally, Vˆ x = V boundary layer region with laminar flow hypothesized to occur in the very thin region near the surface where the velocity gradient (the rate at which the velocity changes with the y-direction near the no-slip boundary) is large. This thin region is called the laminar sublayer. ˆ ∞ , a representative profile If we connect the loci of the points at which Vˆ x reaches V showing the boundary layer growth is obtained. For the case of the boundary layer for flow parallel to a plane surface, the criterion for the type of flow is the local Reynolds number, Rex , which is defined as Rex =

ˆ x V x Vˆ =  

(16.4)

V

V∞ V Edge of Boundary Layer

Laminar

Turbulent Transition

FIGURE 16.7 Velocity profiles for flow along a flat surface.

Viscous Flow

513

where the length scale, x, is the distance along the surface measured from the leading edge. The values of Rex which relate to the type of flow are given for three conditions: (a) For 0 < Rex < 2 × 105 , the boundary layer flow is laminar. (b) For 2 × 105 < Rex < 3 × 10, 6 the boundary layer flow may be either laminar or turbulent. (c) For 3 × 106 < Rex , the boundary layer flow is turbulent. Because there is no precise location where viscous forces suddenly become negligible, the thickness of the boundary layer, normally designated, , is fictitious. The criterion that is normally used is that the boundary layer is considered to extend away from the surface a distance such that Vˆ x = 0.99 Vˆ ∞ or when the velocity is within 1% of the free stream value. With this concept, Prandtl divided the flow regime between the boundary layer, where viscous effects are significant, and the free stream, where the flow may be treated as inviscid. Analytical solutions that give quantitative information for viscous flow effects in the boundary layer have been achieved for certain special cases. Flow parallel to a plane surface is the case usually examined in the first course in fluid mechanics. A detailed study of such boundary layer flows involves mathematical techniques that are beyond the scope of this work. The interested reader may refer to the original work of Blasius (1908) or to Welty et al. (2001) for a discussion of this analysis. For the present, we will accept the results of such analyses. For laminar boundary layer flow along a plane surface, the following results have been achieved: •

The thickness of the boundary layer, , can be evaluated according to =5

•

x 1/2

Rex

(16.5)

The local coefficient of skin friction, Cfx , can be determined from Cfx = 0.664Re−1/2 x

(16.6)

The use of these expressions will be demonstrated in Example 16.2 that follows.

Example 16.2 For the case of air at 300 K (27◦ C) flowing at 15 m/s along a plane surface, determine (a) The boundary layer thickness at a location 20 cm from the leading edge (b) The value of Cfx at this location (c) The total viscous drag exerted by the air on the surface whose dimension is 0.20 m in the flow direction (d) The distance, x, from the leading edge where the transition region would begin (e) The distance from the leading edge where the turbulent boundary layer region would begin

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Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) The surface is infinitely wide. (4) Uniform properties exist at each point considered. (a) We will need the value of the Reynolds number. Table A.13 gives at 300 K,  = 15.69 × 10−6 m/s2 so that Rex =

x Vˆ x (0.20m)(15 m/s) = = 1.912 × 105  15.69 × 10−6 m/s2

Because Rex < 2 × 105 , the boundary layer is laminar. Using Equation 16.5, we determine  as  = 5(20 cm)(1.912 × 105 ) −1/2 = 0.229 cm ⇐ (b) Now that we know Rex , we can use Equation 16.6 to evaluate Cfx as Cfx = (0.664)(1.912 × 105 ) −1/2 = 0.00152 ⇐ With this quantity we can evaluate the local drag force, that is, the drag force associated with a differential element that is located 20 cm from the leading edge. (c) A much more interesting quantity is the total drag force over all of the differential areas between x = 0 and x = 20 cm. This is the information sought in part (c) which is clearly obtained by integration of Equation 16.2. 



Fd =

d Fd = A

Cfx A

Vˆ 2∞ dA 2

or Fd =

Vˆ 2∞ 2



L 0

0.664Re−1/2 Wd x x

where W is the plate width. We may continue the calculation to obtain   L Fd Vˆ 2∞  1/2 −1/2 = (0.664) x dx W 2 Vˆ ∞ 0   Vˆ 2∞  1/2 1/2 = (0.664) 2L 2 Vˆ ∞ =

Vˆ 2∞ −1/2 (1.328)LRe L 2

Thus, with  = 1.1774 kg/m3 at 300 K (Table A.13) Fd (1.1774 kg/m3 )(15 m/s) 2 (1.328)(0.20 m) = W 2 (1.912 × 105 ) 1/2

Viscous Flow

515

or Fd = 0.0805 N/m of width ⇐ W (d) The distance from the leading edge where the transition region begins can be found from Rex  Rex ˆ V∞

x= and for Rex = 2 × 105 , the result is x=

15.69 × 10−6 m2 /s (2 × 105 ) = 0.209 m ⇐ 15 m/s

(e) In a similar fashion, the distance from the leading edge where the turbulent boundary layer region begins will be at Rex = 3 × 106 , and we have x=

15.69 × 10−6 m2 /s (3 × 106 ) = 3.14 m ⇐ 15 m/s

The results of the foregoing example provide some interesting insights as to the magnitude of some of the quantities that are being discussed. For example, at a point 20 cm (approximately 7 3/4 in) from the leading edge, the boundary layer is 2.3-mm (0.091 in) thick. Over the first 20 cm of a 1 m wide plate, the drag force exerted by the air is 0.0805 N (0.358 lbf ). It was also demonstrated that the total drag force could be obtained by evaluating the sum of all local contributions up to a specific position. A more useful way to evaluate the total drag is to use the mean coefficient of skin friction defined as C¯ f =

F /A Vˆ 2 /2

(16.7)

∞

Analysis has given us an expression for Cfx , the local coefficient. The mean coefficient is related to Cfx as F = C¯ f



Vˆ 2∞ A= 2

Cfx A

Vˆ 2∞ dA 2

In general then C¯ f =

1 A

 Cfx dA

(16.8)

A

and for a fixed-plate width, W C¯ f =

1 L



L

Cfx dx 0

(16.9)

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Introduction to Thermal and Fluid Engineering

Now, because we have an expression for Cfx for laminar boundary layer flow, its substitution into Equation 16.9 yields  1 L ¯ Cf = 0.664 Re−1/2 dx x L 0    1/2  1 L  1/2 −1/2  = 0.664 x dx = 1.328 L 0 Vˆ ∞ L Vˆ ∞ or −1/2 Cˆ f = 1.328Re L

(16.10)

The parameter, Re L , is a total Reynolds number with length scale, L, being the dimension of the area of interest along the direction of flow. The local Reynolds number, Rex , contains a length scale, x, which applies at a particular point in the flow direction. For a given length of plate, there is only one value of Re L but an infinite number of values for Rex , one being at the point where x = L or, in magnitude, Rex = Re L . This distinction between the local and mean parameters will be made again as we continue. For turbulent boundary layer flow along a plane surface, these same quantities, , Cfx , and CfL have been determined. However, the techniques involved are less precise than for laminar flow. The reader is again referred to Welty et al. (2001) for the relevant details of this evaluation. The following expressions apply for turbulent boundary layer flow along a plane surface. Recall that boundary layer flow will be turbulent for Rex > 3 × 106 . =

0.376x Re1/5 x

Cfx =

0.0576

CfL =

0.072

Re1/5 x

(16.11)

(16.12)

and Re1/5 x

(16.13)

The remaining question to be considered in this discussion of boundary layer flow along a plane wall is “How do we characterize flow along that portion of the surface where 2 × 105 < Rex < 3 × 106 ?” This is the so-called “transition region” between fully laminar and fully turbulent boundary layer flow. At a location within this region, flow will fluctuate between laminar and turbulent. Near the lower limit (Rex ≈ 2 × 105 ), the flow will be laminar most of the time and occasionally turbulent. Near the upper limit (Rex ≈ 3 × 106 ), the flow will be turbulent most of the time and occasionally laminar.

16.6

Summary

Viscous flow effects have drawn our attention throughout this chapter. We have been particularly interested in the force exerted by a viscous fluid as it flows around or past or solid surface.

Viscous Flow

517

Initially, the Reynolds number criterion for the transition between laminar and turbulent flows was reviewed. The Reynolds number, earlier claimed to be “the most important dimensionless parameter in fluid mechanics,” was encountered throughout the treatment in this chapter. Viscous flow about bluff bodies was discussed with the drag coefficient, C D , and the coefficient of skin friction, C f , defined. Particular attention was paid to a circular cylinder in cross flow. Boundary layer separation effects were discussed and quantified. Approximately half of this chapter has been devoted to two-dimensional, steady, incompressible flow parallel to a plane surface. The classic solution, due to Blasius, was considered in some detail. A less precise but more versatile approach was next discussed and the results of this method were found to agree well with the Blasius results. Information obtained from boundary layer analysis included the boundary layer thickness, (x), and the local and mean skin friction coefficients, Cfx and C f , respectively.

16.7

Problems

Boundary Layers 16.1: If a Reynolds’ experiment were to be performed with a 30-mm pipe, determine the flow velocity that occurs at transition. 16.2: An aircraft flies at a speed of 150 km/h at an altitude of 3000 m. Assuming a transitional Reynolds number of Rex = 5 × 105 and behavior of the boundary layer on the wing surfaces like that on a flat plate, estimate the extent of the laminar boundary layer flow along the wing.

pjwstk|402064|1435600767

16.3: A viscous fluid flows past a flat plate such that the boundary layer thickness at a distance of 1.40 m from the leading edge is 12 mm. Assuming laminar flow throughout, determine the boundary layer thickness at distances of 0.20, 2.0, and 20 m from the leading edge. 16.4: If Vˆ ∞ , the upstream velocity of the flow in Problem 16.3 is 1.5 m/s, determine the kinematic viscosity of the fluid. 16.5: Consider the flow of a viscous fluid across a flat plate. The boundary layer thickness at a distance of 1.25 m from the leading edge is 1.25 cm. Assuming laminar flow, determine the boundary layer thicknesses at distances from the leading edge of (a) 0.25 m, (b) 2.0 m, and (c) 12.5 m. 16.6: In Problem 16.5, if the upstream velocity is Vˆ = 1.625 m/s, determine the kinematic viscosity of the fluid. 16.7: Water flows past a plate with an upstream velocity of Vˆ = 0.25 m/s. Determine the boundary layer thickness at distances of (a) 1.75 cm and (b) 8.75 cm. 16.8: Consider laminar flow over a smooth flat plate with a length of 5 m and a breadth of 3 m. The plate is placed in the air with an upstream velocity of 0.625 m/s and 300 K. Determine the boundary layer thickness and wall shear stress at (a) the center of the plate and (b) the trailing edge of the plate. 16.9: An aircraft flies at a velocity of 640 km/h and an altitude of 3500 m. Suppose that the boundary layers on the wing surfaces approximate those on a flat plate and for a transitional Reynolds number of 5 × 105 , determine the extent of the laminar boundary layer flow along the wings.

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Introduction to Thermal and Fluid Engineering

16.10: A fluid flows across a flat plate at 3 m/s. Determine the thickness and location where the boundary layer becomes turbulent for (a) water at 20◦ C, (b) air at 300 K, and atmospheric pressure, and (c) ethylene glycol at 27◦ C. Fluid Drag 16.11: A two-door hardtop has a drag coefficient of 0.45 at road speeds using a reference area of 2.29 m2 . Determine the horsepower required to overcome the drag at a velocity of 30 m/s. Compare this figure with the case of head- and tail-winds of 6 m/s. 16.12: What diameter plate would have the same drag as the automobile in Problem 16.11? 16.13: The drag coefficient for a smooth sphere is shown in Figure P16.13. Determine the speed at the critical Reynolds number for a 140-mm-diameter sphere in the air. 0.5 0.4

CD

0.3 0.2 0.1 0 103

104

105

106

107

Re = Dν/v FIGURE P16.13

16.14: Using the information provided in Figure P16.13, plot a curve of drag as a function of velocity for a 4.166-cm sphere between velocities of 15 m/s and 110 m/s. 16.15: The lift coefficient is defined as CL =

FL Vˆ 2x A/2

where F L is the lifting force. If the lift coefficient for the automobile in Problem 16.11 is 0.40, determine the lift force at a road speed of 62.5 km/h. 16.16: A baseball having a diameter of 7.163 cm is thrown through the air at sea level and 15◦ C. Determine the drag force on the ball if its velocity is 43 m/s. 16.17: A slender metal rod is in the shape of a cylinder, 16 m high and 12 cm in diameter. When placed in a vertical position, wind at 16 m/s and 20◦ C blows across it. Neglect end effects and determine the bending moment about the base of the cylinder. 16.18: Modern subsonic aircraft have been refined to such an extent that 75% of the parasitic drag (that portion of the total aircraft drag not directly associated with producing lift) can be attributed to friction along the external surfaces. For a typical subsonic jet, the parasitic drag coefficient based on the wing area is 0.011. If the wing area is 225 m2 , determine the friction drag on such an aircraft flying at 312.5 km/h at an altitude of 10,000 m. 16.19: A silo in western Iowa is 3.75 m in diameter and 10-m tall. Wind at 25◦ C is blowing across the silo at 8.75 km/h. Determine the bending moment about the base of the silo.

Viscous Flow

519

16.20: Lubricating oil (≈ SAE 50) is flowing at 20◦ C and 0.75 m/s across a smooth flat plate that is 12.5-cm wide and 42-cm long. Determine the friction drag on one side of the plate. 16.21: A 20.32-cm-diameter spherical balloon travels at 8 m/s at sea level. Determine the speed at an altitude of 3 km to yield the same drag force. 16.22: The frequency of shedding vortices for a cylinder is predicted by the equation   Vˆ x 19.7 f = 0.198 1− d Re where f is the frequency of the vortices shed from one side of the cylinder and Re is the free-stream Reynolds number. Determine (a) the frequency of vortex shedding of a 0.50-cm-diameter wire in a 12.5-km/h wind and (b) the wind velocity required to make the vortex shedding equal to zero. 16.23: A sphere of diameter, d, and density, s , falls at a steady rate through a liquid of ˆ density, , and viscosity, . If the Reynolds number, Re = d V/, is less than unity, show that the speed of fall is gd 2 (s − ) Vˆ = 18 16.24: For a given vehicle, being driven at 35 km/h, 30 HP is needed to overcome aerodynamic drag. Estimate the horsepower at 45 km/h. 16.25: Suppose that a parachute is to be designed to safely land a 120-kg paratrooper as if he had jumped without a parachute from a 3.25-m wall. Assume properties of standard air at sea level and determine the diameter of the parachute. 16.26: A sharp, flat plate that is 3-m wide and 1-m long is parallel to an airstream at 27◦ C flowing at 2.15 m/s. Determine (a) the drag on one side of the plate and (b) the thickness of the boundary layer at the trailing edge of the plate. 16.27: Solve Problem 16.26 for water at the same temperature and velocity. 16.28: A 32-km/h wind blows against an outdoor movie screen that is 18-m wide and 6-m tall. Estimate the wind force on the screen. 16.29: For a frontal area of 0.335 m2 and a drag coefficient of C D = 0.85, determine how much more power is required to pedal a bicycle at 9.50 km/h into a 12.5-km/h headwind than at 9.50 km/h through still air. 16.30: Consider the flow of air at 30 m/s along a flat plate. Determine at what distance from the leading edge transition occurs. 16.31: Estimate the drag force on a 1-m-long radio antenna with an average diameter of 0.50 cm moving at a speed of 37.5 km/h. 16.32: A very wide flat plate 1-m long is immersed in a stream of sea level standard air moving at 1.2 m/s. Determine the drag per unit width of the plate. 16.33: A ship is towing a sonar array that approximates a submerged cylinder 0.325 m in diameter and 9.15-m long with its axis normal to the direction of tow. The tow speed is 15 knots (1 knot = 0.515 m/s). Estimate the horsepower required to tow this cylinder. 16.34: A helium-filled balloon is required to have a sea level terminal ascent velocity of 6 m/s. The helium pressure is 125 kPa and the balloon payload weight (not including the helium) is 280 N. Neglecting payload drag, estimate the proper balloon diameter.

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16.35: A hot-air balloon that is roughly spherical in shape has a volume of 0.5296 m3 and a total weight of 0.0508 N. The outside air temperature is 26◦ C, the temperature within the balloon is 81◦ C, and the pressure is atmospheric. Estimate the rate at which the balloon will rise under steady-state conditions. 16.36: A flagpole standing in sea level standard air is 12 cm in diameter and 15-m high. Estimate the wind-induced bending moment about the base of the flagpole when the wind velocity is 20 knots (1 knot = 0.515 m/s). 16.37: A circular chimney that has a diameter of 2.15 m and a height of 42.5 m is exposed to a sea level storm with winds at 27.5 m/s. Estimate (a) the drag force and (b) the bending moment about the bottom of the chimney. 16.38: A vertical 25 cm diameter piling in seawater is 5-m deep and the current velocity is 2.2 m/s. Estimate the drag on the piling. 16.39: A hydrofoil that is 0.375-m wide and 1.875-m long is placed in water at 30◦ C with a flow of 12.5 m/s. Estimate the boundary layer thickness at the trailing edge of the plate. 16.40: For the data of Problem 16.39, estimate the friction drag for the case of turbulent boundary layer flow over the total surface. 16.41: Solve Problem 16.40 for laminar flow at the leading edge and a transition Reynolds number of 5 × 105 . 16.42: Consider a heavy solid sphere of diameter, d, and density, s suddenly dropped from rest into a fluid of density, , and viscosity, . Assuming that C D is constant, set up a ˆ differential equation and solve for (a) the time history, V(t), (b) the final or terminal ˆ ˆ velocity, V f , and (c) the time required for V to reach 99% of Vˆ f . 16.43: A 3.835-cm-diameter table tennis ball, weighing 0.0247 N, is released from the bottom of a swimming pool. Determine its velocity when it breaks the surface if the pool is 2.5-m deep. 16.44: The ping-pong ball with specifications given in Problem 16.43 can be supported by an air jet from a vacuum cleaner exhaust. For sea level standard air, determine the jet velocity required (Figure P16.44).

FIGURE P16.44

16.45: Assume that a radioactive dust particle approximates a sphere with a specific weight of 25 kN/m3 . Determine how long it will take such a particle to settle to earth from an altitude of 10 km if the particle diameter is (a) 1 m and (b) 10 m.

Viscous Flow

521

16.46: Assuming each to behave as a solid sphere, compare the rise velocity of a 0.3175-cmdiameter air bubble in water to the fall velocity of a 0.3175-cm-diameter water drop in the air. 16.47: Estimate the normal force on a circular sign 2.5 m in diameter during a hurricane wind of 75 km/h. 16.48: A square 15-cm piling is subject to a water flow of 1.75 m/s at a depth of 6.75 m. Determine the bending moment at the bottom of the piling. 16.49: A thin flat plate that is 0.625 m by 1.25 m is immersed in a stream of ethylene glycol at 17◦ C flowing at 6 m/s. The stream is parallel to the short side. Determine the viscous drag. 16.50: Rework Problem 16.49 if the flow of the stream is parallel to the long side. 16.51: As indicated in Figure P16.51, a heavy sphere attached to a string should hang ˆ Neglecting the string drag at angle  when immersed in a stream of velocity, V. and assuming that the fluid is sea level standard air at Vˆ = 40 m/s (a) derive an expression for  as a function of sphere and flow properties and (b) determine  for a steel sphere with a specific gravity of 7.86 and a diameter of 4 cm. θ

V

d, rs

FIGURE P16.51

16.52: A small spherical water drop of diameter 0.005 cm exists in the atmosphere at an elevation of 1600 m. Determine whether the drop will rise or fall if it is in an upwardly flowing thermal of (a) 1.25 m/s, (b) 0.300 m/s, and (c) 0.0300 m/s. 16.53: A pickup truck has a projected area of 10 m2 and a measured drag coefficient, C D = 0.35. Estimate the horsepower required to drive the truck at 35 km/h (a) clean and (b) with a 1 m × 2 m sign installed. 16.54: Determine the terminal velocity of a roughly spherical 180-N rock of specific gravity, SG = 1.93, if the rock falls through (a) air and (b) water. 16.55: A parachutist is to land at sea level at a vertical velocity of 15 km/h. The parachutist and pack weigh 935 N. Determine the diameter of the chute. 16.56: A smooth steel (SG = 7.86) sphere is immersed in a stream of ethanol at 17◦ C moving at 1.5 m/s. Take the diameter as 3 cm and determine (a) the drag in Newtons and (b) the stream velocity that would quadruple the drag. 16.57: Determine the drag on a small circular disk of 0.665-cm diameter moving at 0.0305 m/s through oil with a specific gravity of 0.87. The disk is oriented normal to the upstream velocity. 16.58: An aircraft travels at 225 m/s at an altitude of 8.75 km. It has a smooth wing that is 6.75 m × 48 m. Determine the power required to overcome friction drag.

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17 Flow in Pipes and Pipe Networks

Chapter Objectives •

To introduce the concept of frictional head loss in pipes.

To provide a framework for head loss as a function of Reynolds number, relative roughness, and the Fanning friction factor, f f . • To develop the Hagen-Poiseuille equation for f f in laminar pipe flow. •

To introduce the Moody diagram for evaluating f f in transition flow and in fully turbulent flow. • To employ f f in evaluating flow in single-path applications. •

•

17.1

To use f f in evaluating flow in multiple-path applications.

Introduction

Among the things that everyone takes for granted is the manner in which fluids are transported from one place to another so that certain desirable things can occur. Examples of fluids delivered through pipes, ducts, and other confining conduits include building heating and ventilating systems, automobile cooling systems, water systems in municipalities and homes, gas and oil furnaces, hydraulic systems, petroleum refineries, and cooling channels in integrated circuits. Our purpose, in this chapter, is to bring the accumulated knowledge generated in earlier chapters to bear on the questions related to the delivery of fluids from one place to another through conduits. The term “conduit” applies to any closed-flow passage and the shape can be quite irregular in cross section or quite regular (e.g., square or circular). For purposes of this discussion, the generic term, “pipes,” implying a circular cross section, will be used. Later in the chapter, it will be shown that relationships for pipe flow can be altered in relatively simple fashion to apply to ducts of any other shape, no matter how irregular that shape may be. The examination of pipe flow is generally undertaken for two reasons. First, we wish to determine what operating conditions are necessary to achieve a desired rate of flow through an existing flow system (the analysis problem). The other reason is to specify the configuration of a piping system that will deliver the fluid at a desired rate between two points, a source and a sink, where conditions are known (the design or synthesis problem). Involved in such problems are devices that produce the flow such as pumps and fans and devices that control the flow such as valves and fittings. 523

524

Introduction to Thermal and Fluid Engineering

Pipes may be as large as a crude oil pipeline or as small as the cooling channel in an integrated circuit and flows may be laminar or turbulent. A system may be simple, as with a pipe of a constant cross section connecting a source and a sink, or complex, as with a municipal water system consisting of myriad pipe sizes and lengths along with multiple valves and fittings, where flow may be through numerous pipes in a series, or in parallel, or in any combination. After beginning with the simple case of a single pipe with a constant cross section connecting two locations, we will develop the ideas needed in the analysis of more complex systems.

17.2

Frictional Loss in Pipes

The simplest pipe flow case is one in which the source and the sink are at the same elevation, the pipe connecting them has a constant cross section and the flow is steady. Such a case is shown in Figure 17.1 The first law of thermodynamics for this case, written between the upstream (point 1) and downstream (point 2) ends of the pipe reduces to the form gz1 +

Vˆ 21 Vˆ 2 P1 P2 + + u1 = gz2 + 2 + + u2 2  2 

or, equivalently g(z1 − z2 ) +

Vˆ 21 − Vˆ 22 P1 − P2 + = u2 − u1 2 

(17.1)

Note at this point that all quantities with subscripts (e.g., Vˆ 1 ) are the average values at the designated location. For this specific application with points 1 and 2 at the same elevation, z1 − z2 = 0, and for steady flow through a constant diameter pipe, Vˆ 21 − Vˆ 22 = 0. Equation 17.1 thus reduces to the very simple form that relates the pressure drop as a function of the internal energy increase. P1 − P2 = u2 − u1 

1

Flow

d

L FIGURE 17.1 Single-path pipe flow.

(17.2)

2

Flow in Pipes and Pipe Networks

525

z1 z2

1

×

Flow

×

2

L FIGURE 17.2 Single-path pipe flow with manometers to indicate head at stations 1 and 2.

The internal energy will increase because of frictional effects between the flowing fluid and the pipe wall. Our immediate task is that of evaluating this frictional energy loss that can be related to a pressure drop. If the system shown in Figure 17.1 is modified slightly so that the pressures at points 1 and 2 are indicated by the heights of fluid supported in the vertical tubes at those locations as indicated in Figure 17.2 where the tops of the tubes are open to the atmosphere, the equivalent form of Equation 17.2 is g(z1 − z2 ) = u2 − u1

(17.3)

where the internal energy increase is equivalent to an elevation change or head loss due to friction. The term “head loss,” often used in this context is symbolized by h L . Head loss represents a decrease in the capacity of a flow system to do useful mechanical work, hence the term loss. It can be caused by a number of factors that we will consider in this chapter. At this time, for the case described by Equation 17.3, the head loss is related solely to pipe friction.

17.3

Dimensional Analysis of Pipe Flow

As an initial approach to expressing head loss in the pipe flow condition shown in Figure 17.2, we will use dimensionless analysis. Table 17.1 lists the important variables for this case. With the exception of pipe roughness, the quantities listed are familiar by virtue of earlier use. Pipe roughness, designated e, can be thought of as the characteristic height of projections from the pipe wall; hence, its units of length. The term roughness represents the pipe surface condition. A dimensional analysis performed with this set of variables, with the core group comˆ d, and , proceeds as follows: posed of V, The Buckingham Pi theorem gives i = n−r = 7−3 = 4

526

Introduction to Thermal and Fluid Engineering TABLE 17.1

Significant Variables for Dimensional Analysis of Pipe Fow Variable

Symbol

Dimensions

P Vˆ

M/L T 2

Pressure loss Velocity Pipe diameter Pipe length Pipe roughness Fluid density Fluid viscosity

d L e  

L/T L L L M/L 3 M/L T

which shows that there will be four pi groups. They are formed as follows: 1 = Vˆ a d b c P 2 = Vˆ a d b c L 3 = Vˆ a d b c e 4 = Vˆ a d b c  Proceeding as discussed in Chapter 15, we achieve the following forms for the pi groups: 1 =

P Vˆ 2

L d e 3 = d

2 =

and 4 =

 ˆ d V

The first pi group is the Euler number. It contains the frictional pressure drop in the form represented by the left side of Equation 17.2. This term can be expressed in terms of head loss by replacing P/ by its equivalent, gh L , from Equation 17.3. Thus, an equivalent form for 1 is 1 =

gh L Vˆ 2

The parameters 2 and 3 are clearly dimensionless because they are ratios of quantities having units of length. The ratio, 3 = e/d is commonly referred to as the relative roughness. The fourth pi group is the reciprocal of the Reynolds number, Red . In functional form, the pi groups can be written as   hL L e = f , , Red (17.4) d d Vˆ 2 /g Experience has shown that the head loss in fully developed pipe flow is directly proportional to the ratio, L/d. This dimensionless ratio is, thus, not included as a part of the function, f ,

Flow in Pipes and Pipe Networks

527

and Equation 17.4 becomes modified as  hL L e = f , Red d d Vˆ 2 /g The function f

e d

(17.5)

 , Red

which varies with relative roughness and Reynolds number is yet to be evaluated. The head loss expression, in its general operational form, is a modified version of Equation 17.5 and this form is L Vˆ 2 hL = 2 f f (17.6) d g where the factor 2 is associated with f f , the Fanning friction factor. In a variation of Equation 17.5, often seen, we write h L = fd

L Vˆ 2 d 2g

(17.7)

where f d is the Darcy friction factor. It is obvious that f d = 4 f f . The constant, 2, in Equation 17.6 is conventionally associated with the head loss expression incorporating the Fanning friction factor. In this form f f = C f , the coefficient of skin friction, which was introduced in Chapter 16. Equation 16.6, by convention, is written with the 2 in the denominator so that the right-hand side has the form of kinetic energy. In this book, the Fanning friction factor will be used exclusively. Our immediate task is to determine suitable relations for f f from both theory and experimental data.

17.4

Fully Developed Flow

When a fluid such as water enters a pipe, its velocity profile will be altered as viscous effects propagate from the pipe wall into the fluid. The entering profile may be square (this is generally referred to as slug flow) and will be altered owing to the effect of viscosity as the fluid proceeds down its flow path. Figure 17.3 shows the propagation of the velocity profiles between the entrance at x = 0 to the distance, L e , where viscous flow effects have proceeded to the center of the pipe. If the flow remains laminar (recall that this will be true for Red < 2300), the fully developed velocity profile will be parabolic. The fully developed condition will occur at some distance, L e , downstream from the entrance. The distance, L e , is designated as the entrance length. For x < L e the velocity profile will change with the distance, x. In this entry region there will be two regimes of flow. The region affected by the presence of the pipe wall is called the viscous region. This is where velocity gradients exist, that is, ∂Vˆ x /∂r =  0. Near the center of the pipe, the velocity profile remains flat, that is ∂Vˆ x /∂r = 0. This is called the core flow region. For x > L e , the viscous region completely fills the pipe and the velocity profiles will no longer vary with x. This condition is referred to as fully developed flow. Mathematically, fully developed flow requires that ∂Vˆ x /∂x be zero. We should recall from the discussion in Chapter 13 that conservation of mass requires that the flow rate be constant for all values of x. Because the velocity near the wall is decreasing,

528

Introduction to Thermal and Fluid Engineering Le r Flow

x

Slug Flow Profile

Developing Profile

Fully developed Profile

FIGURE 17.3 Velocity profiles for flow in pipes or tubes.

ˆ max = 2V ˆ av at the the rate of flow in the core must increase, eventually reaching a value V centerline for fully developed laminar flow. The concepts of entrance length and fully developed flow are also valid when the flow is turbulent; the velocity profiles will be different, however, and the fully developed profile will not be parabolic.

17.5

Friction Factors for Fully Developed Flow

17.5.1 Laminar Flow The case of fully developed laminar flow in a circular flow passage is amenable to a straightforward analytical description. The interested reader may refer to a variety of fluid mechanics texts for a detailed discussion [Fox and McDonald (1985), Welty et al. (2008), and White (1999)]. The result of such analyses is the classic Hagen1 -Poiseuille2 equation, which is stated here without proof −

dP Vˆ = 32 2 dx d

(17.8)

This equation is easily solved by separation of the variables and integration  −

P P0

d P = 32

 L Vˆ dx d2 0

or L P = P0 − P = 32Vˆ d

(17.9)

The ratio P/ is the internal energy change as expressed by Equation 17.2. We can thus write P  L = gh L = 32 Vˆ   d

(17.10)

1 G.

H. Hagen (1797–1884) was a German hydraulic engineer who developed the Hagen-Poiseuille law independent of J. L. M. Poiseuille and directed construction of dikes, harbor installations, and dune fortifications 2 J.

L. M. Poiseuille (1797–1869) was a French physiologist and physicist who studied the physiology of arterial circulation, invented improved methods for measuring blood pressure, and discovered the Hagen-Poiseuille law independently of G. H. Hagen.

Flow in Pipes and Pipe Networks

529

which can be related to Equation 17.6 to introduce f f . Doing so, we have h L = 32

 ˆL Vˆ 2 L V = 2ff g d g d

The resulting expression for f f in laminar flow is ff =

16 16 = ˆ Red Vd/

(17.11)

Equation 17.11 verifies the dimensional analysis result that the friction head loss is functionally related to the Reynolds number. This result has also been verified experimentally. The absence of any dependency on pipe roughness in laminar flow has also been experimentally verified. While this may seem counterintuitive, the physical explanation is that any flow disruption due to wall roughness is dampened out by viscous effects. Recall that for values of Red in pipe flow less than 2300, viscous effects are predominant over inertial effects. This discussion suggests that roughness may be a significant factor with turbulent flow. This case will be examined next. 17.5.2 Turbulent Flow In the case of turbulent flow, there is no neat and simple relationship such as the HagenPoiseuille equation to evaluate the frictional pressure drop. For values of Red > 2300, a range that encompasses both the transition and fully turbulent regimes (we will identify these regimes shortly), the relationships for f f have been obtained by incorporating both analytical approaches, using turbulent flow analysis and empiricism. For details, the interested reader is, again, referred to the available fluid mechanics texts such as Welty et al. (2001). The following relationships and range of application for f f in pipe flow are now summarized: •

Laminar flow, Red < 2300 (repeated here in order to show a complete catalog) ff =

•

16 Red

(17.11)

Turbulent flow, Red > 2300, hydraulically smooth pipe

 1  = 4.0 log10 [Red f f )] − 0.40 ff  • Turbulent flow, Red > 2300, rough pipe, (d/e)/Red f f < 0.01 1 d  = 4.0 log10 + 2.28 e ff  • Transition flow, Red > 2300, (d/e)/Red f f > 0.01  1 d d/e  = 4.0 log10 + 2.28 − 4.0 log10 4.67  +1 e ff Red f f

(17.12)

(17.13)

(17.14)

These four equations provide the basis for a graphical representation of f f as a function of Red , which is a standard tool for pipe flow analysis.

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Introduction to Thermal and Fluid Engineering

The range of applications for Equations 17.12 through 17.14 includes the terms “smooth” and “rough.” The accepted criterion for the term “hydraulically smooth” is when the parameter d/e  < 0.01 Red f f

(17.15)

One may observe that for fully turbulent flow in smooth pipes, according to Equation 17.12, the Fanning friction factor is not functionally dependent on the relative roughness and is a function only of Red . Similarly, for rough pipes in fully turbulent flow, f f is solely a function of relative roughness.

17.6

Friction Factor and Head Loss Determination for Pipe Flow

17.6.1 Pipe Friction Factor Figure 17.4 plots f f as a function of Red , with separate lines for a range of e/d values in the turbulent regime. This Figure is similar to a plot presented by Moody (1944) and such plots are frequently referred to as Moody diagrams or Moody charts. One may note the three regimes shown in Figure 17.4. They are •

The laminar flow regime where f f = f (Red )

•

The fully turbulent regime where ff = f

•

The transition regime where

e  d

 e f f = f Red , d

The remaining piece of information necessary for the use of Figure 17.4 is the value for the relative roughness, e. Some typical values of the pipe roughness for common construction materials are listed in Table 17.2. One should be aware that these values for pipe roughness are reasonable approximations when pipe surfaces are free from corrosion or foreign material. The values listed in Table 17.2 are used, in general, along with the friction factor plot, to solve pipe flow analysis or design problems. 17.6.2 Head Loss Due to Fittings and Valves Additional head losses will occur in a piping system due to the presence of valves, elbows, and a variety of other fittings that involve changes in flow direction, changes in the size of the flow passages or control of the flow rate and/or the pressure. Head losses associated with such elements are functions of geometry, roughness, and the Reynolds number. The Reynolds number dependence being relatively small, the head loss associated with valves and fittings can be reasonably approximated as hL =

P Vˆ 2 =K g 2g

where the factor, K , has a specific value for a given fitting or valve.

(17.16)

Flow in Pipes and Pipe Networks

531

0.025 Laminar Flow Complete Turbulence Rough Pipe

0.02

0.05 0.04

0.015

0.03 0.025 0.02

0.01 0.008 0.006

0.009 0.008

0.004 0.003

0.007 Uncertain Region

0.006

0.002 0.0015

0.005

0.001 0.0008 0.0006

0.004

0.0004

Relative Roughness, e/d

Fanning Fraction Factor, ff

0.015 0.01

0.0002 0.0001

0.003

5 × 10–5 3 × 10–5 2 × 10–5 5 × 10–6 1 × 10–5

0.0025 0.002

Smooth Pipe (e = O) 68

103

2 34 68

2 34 68

104

2 34 68

105

2 34 68

106

107

2 34 68

108

 Reynolds Number = ρVd/μ FIGURE 17.4 The Fanning friction factor as a function of the Reynolds number and relative pipe roughness.

TABLE 17.2

Values of Pipe Roughness, e, for Several Materials Material Drawn tubing Commercial steel Wrought iron Asphalted cast iron Galvanized iron Cast iron Wood stave Concrete Riveted steel

e, m ×104 0.0152 0.457 0.457 1.22 1.52 2.59 3.0–9.1 30.5 9.1–91.4

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Introduction to Thermal and Fluid Engineering

Another approach to accounting for the head loss due to minor losses is the concept of an equivalent length. The idea here is to determine the length of straight pipe that would involve the same head loss as a particular valve or fitting. Thus, a single expression of the form hL = 2 f f

L eq Vˆ 2 d 2g

(17.17)

can be used for a given system where L eq accounts for the pipe itself plus all of the minor losses present in the form of valves and fittings. In comparing Equations 17.16 and 17.17 it is clear that the factor K is equivalent to 4 f f L eq /d. It appears that Equation 17.17 is Reynolds number dependent because it includes f f , but this is not the case. The assumption is made that, in both cases, the Reynolds number is so large that the flow is fully turbulent. In this case, the friction factor will depend only on the pipe roughness. Some values for K and L eq /d are listed in Table 17.3. An additional head loss expression that will prove useful is associated with a sudden expansion. When the cross-sectional area of a pipe changes abruptly, there is an associated head loss that can be determined from   Vˆ 2 A1 2 hL = 1 1 − (17.18) 2g A2 where the subscripts 1 and 2 refer to upstream and downstream conditions, respectively. The term   A1 2 1− A2 TABLE 17.3

Friction Coefficients and Equivalent Lengths for Pipe Fittings Fitting Valves

Elbows

Bends Tees

Union

Globe, fully open Angle, fully open Gate, fully open Gate, 3/4 open Gate, 1/2 open Gate, 1/4 open Ball, fully open Ball, 2/3 open Ball, 1/3 open Regular, 90◦ , flanged Regular, 90◦ , threaded Long radius, 90◦ , flanged Long radius, 90◦ , threaded Regular, 45◦ , threaded Long radius, 45◦ , flanged 180◦ , threaded 180◦ , flanged Straight through flanged Straight through threaded Side outlet flanged Side outlet threaded Threaded

K

Leq , m

10 2 0.15 0.26 2.1 17 0.05 5.5 210 0.3 1.5 0.2 0.7 0.4 0.2 1.5 0.2 0.2 0.9 1 2 0.08

470 90 7 7 12 800 2 260 10,000 14 70 9 30 19 9 70 9 9 40 50 90 4

Flow in Pipes and Pipe Networks

533

(a)

(b)

(c)

(d)

FIGURE 17.5 Entrance shapes for four typical inlet geometries.

represents the loss coefficient, K , in this case. The value of K for a sudden expansion is seen to vary when A2 is very large, so that as A1 /A2 −→ 0, K = 1, and as A1 /A2 −→ 1, K = 0. In the reverse case, that of a sudden contraction in the flow passage, the loss coefficient will have a range in values that will vary with the shape of the inlet. Figure 17.5 illustrates four representative inlet shapes for the case of flow from a large tank or reservoir into a circular pipe (A2 /A1 ≈ 0). In the case of flow from a larger pipe to a smaller one, with 0 < A2 /A1 < 1, the corresponding value of K will vary from 0.5 when A2 /A1 ≈ 0, as in Figure 17.5b, to 0 when A2 /A1 ≈ 1. This variation of K as a function of A2 /A1 is shown in Figure 17.6. The information presented in Table 17.3 and in Figures 17.5 and 17.6 is representative, but by no means exhaustive, of the various cases for which loss coefficients in piping systems may be determined. More detailed and complete information of this kind is available in numerous references such as Munson et al. (1994). 0.6 A1

A2

hL = KL

V22 2g

0.4 KL 0.2

0

0

0.2

0.4

0.6 A2/A1

FIGURE 17.6 Loss coefficient for a sudden contraction.

0.8

1.0

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Introduction to Thermal and Fluid Engineering

17.6.3 Noncircular Flow Passages All equations used this far have included the diameter, d, implying a circular cross section. When the flow passage is noncircular, these same expressions can be used with d replaced by the equivalent diameter, deq . The equivalent diameter, sometimes referred to as the hydraulic diameter, is determined according to deq = 4

cross-sectional area wetted perimeter

(17.19)

It is a simple matter to verify that deq = d for a circular flow passage. For other shapes, deq may have values which one would not predict intuitively. For example, in the case of an annular flow passage between two circular pipes (inner pipe with di and outer pipe with do ), application of Equation 17.19 yields   2 do − di2 (do + di )(do − di ) 4 deq = 4 = = do − di (do + di ) do + di 17.6.4 Single-Path Pipe Systems Examples 17.1 through 17.3 that follow, illustrate the application of the material presented in earlier sections to the analysis and design of single-path piping systems.

Example 17.1 In one section of the Alaska pipeline, (Figure 17.7), crude oil at 55◦ C ( =

864 kg/m3 and  = 4.32 × 10−3 kg/m-s) is pumped at a rate of 3.4 m3 /s through a 1.22-mdiameter steel pipe. Determine the pumping power required to produce this flow over a length of 850 miles (1370 km) between two pumping stations.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) There are no entrance and exit losses to the pipe. (4) There are no losses due to bends, pipes, and fittings. (5) Heat transfer and potential energy changes are negligible. (6) The pressures at the upstream and downstream end are equal. (7) The pump is considered as exterior to the pipe section. (8) Uniform properties exist at each cross section considered. 1

2

3.4 m3/s

d = 1.22 m

1370 km FIGURE 17.7 Piping configuration for Example 17.1.

Flow in Pipes and Pipe Networks

535

This is an analysis problem because the size, length, and configuration of the pipe are established. The pumping power is related to the other quantities by the first law of thermodynamics. For the 1370-km-long control volume between the upstream location (point 1) and the downstream location (point 2), the steady-flow statement of the first law is  Vˆ 21 − Vˆ 22 P1 − P2 ˙ ˙ Qcv − Wcv + m ˙ g(z1 − z2 ) + + + (u1 − u2 ) = 0 2  For this case, the terms are evaluated as: The heat transfer and potential energy changes are assumed negligible ˙ cv = 0 Q •

and

g(z1 − z2 ) = 0

For steady flow through a constant diameter pipe, Vˆ 1 = Vˆ 2 Vˆ 21 − Vˆ 22 =0 2

•

For upstream and downstream conditions at equal pressures P1 − P2 =0 

•

There is a relationship between the head loss and the change in specific internal energy

L Vˆ 2 u2 − u1 = gh L = g 2 f f d g

and the first law relationship reduces to   L ˙ cv = m −W ˙ 2 f f Vˆ 2 d For crude oil at 55◦ C, the following properties apply:  = 864 kg/m3

and

 = 4.32 × 10−3 N-s/m2

The Reynolds number is evaluated as Red = =

ˆ ˙ ˙ d V d V V = =4  A d 4(3.4 m3 /s)(864 kg/m3 ) (1.22 m)(4.32 × 10−3 N-s/m2 )



1 N-s2 kg-m

= 709,000

and the flow is turbulent. With the value of the pipe roughness, e = 0.457 × 10−4 m taken from Table 17.2 for commercial steel, we have the relative roughness e 0.457 × 10−4 m = = 3.75 × 10−5 d 1.22 m

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Introduction to Thermal and Fluid Engineering

and the value of f f can be read from Figure 17.4 f f = 0.0033 Now with ˙ ˙ V 4V 4(3.4 m3 /s) Vˆ = = = = 2.909 m/s A d 2 (1.22 m) 2 we can now complete the solution   L ˆ2 ˙ −Wcv = m ˙ 2ff V d     1.37 × 106 m = (3.4 m3 /s)(864 kg/m3 ) 2(0.0033) (2.909 m/s) 2 1.22 m or

˙ cv = −18.42 × W

kg-m 106 s2

2



1 N-s2 kg-m

W = −184.2 MW

The negative value indicates that work is being done on the fluid.

Example 17.2 A pump delivers 0.03 m3 /s of water through a 15.5-cm-diameter commer-

cial steel pipeline as displayed in Figure 17.8, which shows two long radius 45◦ elbows. For a pump discharge pressure of 810 kPa, determine the pressure at the pipe exit. 45° Elbow

0.1 km

Storage Tank B

Pump

A

45° Elbow

0.1 km

0.1 km FIGURE 17.8 Pumping configuration for Example 17.2.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) There are no entrance and exit losses to the pipe. (4) There are no losses due bends, pipes, and fittings. (5) Heat transfer is negligible. (6) The pump is treated as exterior to the pipe section. (7) Uniform properties exist at each cross section considered.

Flow in Pipes and Pipe Networks

537

The analysis will be based on a control volume that encloses the pipe and extends from the pump discharge (point A) to the entrance of the storage tank (point B). For this control volume, the energy equation takes the form g(z B − z A) +

PB − PA + uB − u A = 0 

which may be written as PA − PB = g(z B − ZA) + gh L  where PB is the unknown being sought. The potential energy term is evaluated as g(z B − z A) = (9.81 m/s2 )(100 sin 45◦ m) = 694 m2 /s2 The head-loss term includes the frictional effect of flow through 300 m of pipe plus the loss due to two 45◦ elbows. To evaluate the pipe friction, we must determine the Reynolds number. Using the properties of water at 20◦ C from Table A.14 in Appendix A, we find  = 998.0 kg/m3

and

 = 1.012 × 10−6 m2 /s

Then we have Red =

˙ d Vˆ 4V 4(0.03 m3 /s) = = = 243,500  d (1.012 × 10−6 m/s2 )(0.155 m)

which is well into the turbulent range. The relative roughness is evaluated as e 0.457 × 10−4 m = = 2.95 × 10−4 d 0.155 m With this information, the Fanning friction factor is determined from Figure 17.4 as f f = 0.0044 The determination of the minor losses involves Equation 17.19. For a 45◦ elbow (flanged) the value of K from Table 17.3 is 0.2. Then, the head loss for two elbows and the pipe friction will be L ˆ2 V + 2K Vˆ 2 d   L = 2Vˆ 2 f f + K d

gh L = 2 f f



4(0.30 m3 /s) =2 (0.155 m) 2 = 44.1 m2 /s2

2     300 m (0.0044) + 0.2 0.155 m

538

Introduction to Thermal and Fluid Engineering

Returning to our energy equation expression, we now have PA − PB = 694 m2 /s2 + 44.1 m2 /s2 = 738.1 m2 /s2  Finally, the pipe exit pressure is evaluated as PB = PA − (738.1 m2 /s2 ) = 810,000 N/m2 −

(998.1 kg/m3 )(738.1 m2 /s2 ) 1 kg-m/N-s2

= 810,000 N/m2 − 736,700 N/m2 = 73,300 N/m2 = 73.3 kPa ⇐

Example 17.3 Water at a temperature of 20◦ C is pumped through the system shown in Figure 17.9 by a pump that develops 164 kW. The discharge pressure is 279 kPa (gage). Determine the volumetric flow rate of the water.

(1) 60 m

Reservoir

2 Storage × Tank

60 m 25 m

Pump

30 m ×

Rounded Entrance K = 0.2 FIGURE 17.9 Piping configuration for Example 17.3.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) The pump is considered as exterior to the pipe section. (4) Uniform properties exist at each cross section considered. This is a variation of the analysis problem where the unknown quantity is the flow rate. A convenient choice for the control volume is between point 1 at the surface of the reservoir and point 2 where the storage tank pressure is known. The first law of thermodynamics

Flow in Pipes and Pipe Networks

539

between these points is 

Vˆ 22 − Vˆ 21 P − P 2 1 ˙ cv − m −W ˙ g(z2 − z1 ) + + + u2 − u1 = 0 2  A term-by-term evaluation with  = 998.1 kg/m3 and  = 1.012 × 10−6 m/s2 (Table A.14) yields ˙ cv = −(−164 kW) = 164 kW −W ˙ = 998.1V ˙ kg/s m ˙ = V g(z2 − z1 ) = (9.81 m/s2 )(25 m − 60 m) = −343.4 m2 /s2   2 ˙ Vˆ 22 − Vˆ 21 1 4V ˙ 2 m2 /s2 = − 0 = 507V 2 2 (0.20 m) 2 P2 − P1 279 kPa + Patm − Patm = = 280 m2 /s2  998.1 kg/m3 and with two 90◦ -flanged elbows where K = 0.30 u2 − u1 = gh L L ˆ 2  Vˆ 2 V + K d 2     ˙ 2 L K 4V = 2ff + d 2 d 2 = 2ff

  2 ˙ 115 m 1 4V = 2ff + (0.2 + 0.6) 0.2 m 2 (0.2 m) 2 ˙ 2 ) m2 /s2 = (1150 f f + 0.40)(1013V ˙ 2 m2 /s2 = (1.165 × 106 f f + 405) V The u2 − u1 term cannot be evaluated further at this time because f f is a function of Red which, in turn, is a function of the flow rate. Dividing both sides of the energy equation by m ˙ and substituting terms that have been evaluated in the foregoing itemized term-by-term evaluation we obtain 164 × 103 W ˙ 2 m2 /s2 + 280 m2 /s2 = −343.4 m2 /s2 + 507 V ˙ 998.1V ˙ 2 m2 /s2 + (1.165 × 106 f f + 405) V and we see that all terms have the same units (m2 /s2 ). When we perform some algebra, this expression simplifies to ˙ + (1.165 × 106 f f + 912) V ˙3 164 = −63.4V

(a)

540

Introduction to Thermal and Fluid Engineering

˙ is in m3 /s and f f is a function of the Reynolds number where V Red =

˙ d Vˆ 4V =  d

Using properties of water at 20◦ C, the Reynolds number becomes Red =

˙ 4V (0.20 m)(1.012 ×

10−6

2

m/s )

˙ = 6.292 × 106 V

(b)

To complete the solution, an additional piece of independent information relating f f ˙ is needed. Such information is provided by the friction factor plot given in Figure and V 17.4. Because this second piece of information is given in graphical form, an iterative, or trial-and-error solution is required. The solution proceeds as follows: 1. Assume a value for f f . 2. With this value for f f , use Figure 17.4 to find the corresponding value for Red . ˙ 3. Having Red , solve Equation (b) for V.

˙ 4. Use the assumed value for f f , solve Equation (a) for V. ˙ as calculated in steps 3 and 4 agree, the desired result has been 5. If the values for V found and you may stop. ˙ do not agree, adjust the value of f f and go back to step 2. 6. If the two values of V ˙ agree. 7. Repeat the procedure until the values of V To illustrate this procedure, we will first assume the flow to be fully turbulent. For such a case, f f is a function only of e/d, which for this problem is e 0.457 × 10−4 m = = 0.000229 d 0.20 m giving f f = 0.0037. ˙ from Equation (a) to obtain To start the iterative process, we go to step 4 and evaluate V ˙ = 0.328 m3 /s V ˙ Red = 2.013 × 106 and, using Figure Then using Equation (b) we find, for this value of V, 17.4, we find that for e/d = 0.000229 and Red = 2.013 × 106 , the value of f f ∼ = 0.0040, a slightly different value than we assumed. ˙ from Equation (a), we get Then, using f f = 0.0037 and calculating V ˙ = 0.331 m3 /s V for which, from Equation (b) Red = 2.082 × 106 A check of Figure 17.4 indicates that these values are in agreement so that the solution may be considered as completed. The flow rate of water delivered by the pump is 0.331 m3 /s. We now turn to three design examples. •

Design Example 9 (Section 17.7) involves the sizing of a piping system that employs commercial steel pipe, contains 90◦ elbows, and a hydraulic turbine.

Flow in Pipes and Pipe Networks

541

•

Design Example 10 (Section 17.8) considers an irrigation system in which the size of PVC pipe is to be obtained. The system also contains a pump.

•

Design Example 11 (Section 17.9) pertains to the transport of crude oil from an unloading area to a refinery located several kilometers away. A heater is placed at a strategic location to modify the flow properties of the oil and the heater output for a prescribed pumping head is sought.

17.7

Design Example 9

In the system shown in Figure 17.10, water at 16◦ C flows from a reservoir through a piping system that includes a hydraulic turbine. The discharge pressure is atmospheric, the water flow rate is 1.6 m3 /s, and 75 kW is generated by the turbine. Determine the size of commercial steel pipe required.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Minor losses for two 90◦ bends are to be included. (4) The entrance to the pipe from the reservoir is assumed sharp-edged. (5) Heat transfer is negligible. (6) The pressures at the upstream and downstream ends are equal. (7) The turbine is considered as exterior to the pipe section. (8) Uniform properties exist at each cross section considered. (9) Each of the 90◦ elbows is flanged. We apply the energy equation between the reservoir surface (point 1) and the pipe discharge (point 2) and obtain  Vˆ 22 − Vˆ 21 P − P 2 1 ˙ cv = m −W ˙ g(z2 − z1 ) + + + u2 − u1 = 0 2  1 10 m

20 m

30 m 30 m

20 m FIGURE 17.10 Piping configuration for Design Example 9.

Turbine

2

542

Introduction to Thermal and Fluid Engineering

A term-by-term evaluation yields the following: ˙ cv = −75 W −W ˙ = (1000 kg/m3 )(1.6 m3 /s) = 1600 kg/s m ˙ = V g(z2 − z1 ) = (9.81 m/s2 )(0 m − 40 m) = −392.4 m2 /s2  ˙ 2  2 Vˆ 22 − Vˆ 21 Vˆ 22 − 0 1 4V 1 4(1.6 m3 /s) 2.08 = = = = 4 m2 /s2 2 2 2 2 2 d 2 d d where d is in meters. Because P1 = P2 = Patm P2 − P1 =0  and

   L ˆ 2  Vˆ 2 L K ˆ2 u2 − u1 = gh L = 2 f f V + = 2ff + V K d 2 d 2

Minor losses will include the effects of two 90◦ elbows and the entrance from the reservoir into the pipe section that will be assumed to be sharp edged. With these coefficients chosen from Table 17.3 and Figure 17.6, the head loss term becomes u2 − u1 = gh L   100 m 1 = 2ff + (0.3 + 0.3 + 0.5) Vˆ 2 d 2 which results in

  ff gh L = 200 + 0.55 Vˆ 2 d

With these terms substituted into the first law expression, we have  2.08 −75 kW = (1600 kg/s) −392.4 m2 /s2 + 4 m2 /s2 d    ff + 200 + 0.55 Vˆ 2 m2 /s2 d Dividing by m ˙ and further simplifying, we have    ˙ 2 ff 2.08 4V −46.9 = −392.4 + 4 + 200 + 0.55 d d d 2 where all of the terms have units of m2 /s2 . In its simplest form, the energy equation is 345.5 =

ff 4.36 + 830 5 4 d d

(a)

The Fanning friction factor and pipe diameter are related by the friction factor plot in Figure 17.4 where the Reynolds number has the form Red =

˙ d Vˆ 4V =  d

Flow in Pipes and Pipe Networks

543

Using properties of water at 16◦ C,  = 1000 kg/m3

 = 1.310 × 10−6 m/s2

and

the Reynolds number becomes Red =

4(1.6 m3 /s) d(1.310 × 10−6 m/s2 )

or Red =

1.555 × 106 d

(b)

As was the case in Example 7.2, the solution to this problem will require trial and error because the two conditions to be satisfied, Equation (a) and Figure 17.4, are not in forms that allow d to be determined explicity. The procedure to evaluate d will be as follows: 1. Assume a value for f f,old . 2. With this value, solve Equation (a) for d. 3. Use Equation (b) to evaluate Red . 4. Evaluate e/d (for commercial steel, e = 0.457 × 10−4 m). 5. Using Red and e/d, read f f,new from Figure 17.4. 6. If f f,new = f f,old , you may stop because d has the value calculated in step 2. 7. If f f,new =  f f,old , adjust the value of f f,old ( f f,old = f f,new is a possible choice) and go back to step 2. 8. Repeat the procedure until the values of f f,new and f f,old agree. Employing this procedure, we obtain the following results: 1. Assume f f,old = 0.003. 2. Equation (a) becomes 1 345.5 = 4 d



2.49 4.36 + d



from which, by trial and error, we find d = 0.416 m 3. Using Equation (b) we obtain Red = 3.738 × 106 . 4. The relative roughness is e 0.457 × 10−4 m = = 1.099 × 10−4 d 0.416 m 5. From Figure 17.4, we obtain f f,new = 0.00315. 6. Because f f,new =  f f,old , we let f f,old = f f,new = 0.00315 and return to step 2. Repeating this procedure with f f = 0.00315, the value of d is 0.418 m, which is the solution to our problem. The required pipe diameter is 0.418 m.

544

Introduction to Thermal and Fluid Engineering

Because pipes are made in standard sizes, in an actual design one would select a pipe whose inside diameter is as close as possible to this value. Generally, the deviation from operational specifications will be quite small when such a choice is made.

17.8

Design Example 10

A wheat farmer in eastern Oregon plans to use a spray irrigation system for one of his fields. His system will utilize water from a large holding pond that is 38 m below the intended sprayer location. The irrigation system will require that 90 gal/min (0.00568 m3 /s) be delivered to the sprayer at 34 psig (234.4 kPa) g . The pump that is available is capable of providing the needed flow rate at a discharge pressure of 90 psig (620.6 kPa gage). The pump will be sited at a location that is 67 m from the sprayer. Neglecting any minor losses, determine the minimum diameter of PVC pipe that will be suitable for this application.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The PVC pipe may be considered hydraulically smooth. (3) Minor losses are negligible. (4) Water properties are to be taken at 50◦ C. Strategy 1. The system to be evaluated extends from the pump discharge to the sprayer intake and is composed of PVC pipe acting as a conduit for the water flow. 2. The energy equation (First Law of Thermodynamics), Equation 17.1 g(z1 − z2 ) +

Vˆ 21 − Vˆ 22 P1 − P2 + = u2 − u1 2 

(17.1)

must be satisfied. It is clear that •

The required flow rate is specified as 0.00568 m3 /s.

•

The velocity will be unknown until the pipe diameter is determined.

3. The frictional head loss between the pipe entrance and discharge must satisfy the Fanning head loss expression of Equation 17.6 hL = 2 f f

L Vˆ 2 d g

where d = diameter of the pipe which is the parameter sought Vˆ = velocity that is directly proportional to the diameter

(17.6)

Flow in Pipes and Pipe Networks

545

f f = Fanning friction factor that is determined using Figure 17.4. Note that the independent variable is the Reynolds number which includes both diameter and velocity 4. With the energy equation involving the unknowns, Vˆ and d, in complex ways ( f f , d, and Vˆ are all related graphically in Figure 17.4), it is clear that a trial-anderror procedure must be employed. Now that the strategy has been established, we proceed to the solution of the problem. The energy equation applied to a control volume, which includes the pipe with inflow at the pump discharge (section 1) and outflow at the sprayer inlet (section 2), takes the form h L = (z1 − z2 ) +

Vˆ 21 − Vˆ 22 P1 − P2 u2 − u1 + = 2g g g

Note that there is no heat transfer and no shaft work done on or by the control volume and that for water at 50◦ C, Table A.14 provides  = 988.1 kg/m3

 = 0.5435 × 10−6 m/s2

and

Evaluating each term in this relationship, we have z1 − z2 = −38 m Vˆ 21 − Vˆ 22 =0 2g P1 − P2 620.6 kPa − 234.4 kPa = = 39.84 m g (988.1 kg/m3 )(9.81 m/s2 ) Moreover, with ˙ 4V 4 Vˆ = = d 2 



0.00568 m3 /s d2

 =

7.232 × 10−3 m/s d2

the head loss term may be evaluated as hL =

2ff L ˆ2 V g d 2ff



2 7.232 × 10−3 m/s d2

=

67 m 2 d 9.81 m/s

=

7.144 × 10−4 ff m d5

The energy equation may be evaluated as −38 m + 39.84 m =

7.144 × 10−4 ff d5

or 1.84 m =

7.144 × 10−4 ff m d5

546

Introduction to Thermal and Fluid Engineering

As discussed earlier, f f is a function of the Reynolds number that is Red = =

d Vˆ d =  



7.232 × 10−3 m/s d2

(7.232 × 10−3 /d 2 ) m/s 0.5435 ×

10−6

=

2

m/s



13,300 d

We are now ready to proceed with our trial-and-error solution. The computations are summarized in Table 17.4. TABLE 17.4

Computations to Establish the PVC Pipe Diameter Assume d, m Red ff 7.1441 × 10−4 f f /d 5 , m−5

0.08 169,290 0.0042 0.017

0.065 208,350 0.0039 2.403

0.068 199,160 0.0040 1.967

0.0689 193,000 0.0040 1.840

The required conditions are met for d = 0.0689 m. Thus, our required pipe diameter is 0.0689 m

or

2.71 in ⇐

It is frequently necessary to transport crude oil from an unloading area to a refinery. In such a case, a system is selected and the system output is calculated. This design problem considers the capacity of an available pipeline as well as the economic advantage of heating the oil to increase the capacity.

17.9

Design Example 11

An oil refinery is located 11 km from a marine terminal where oil tankers are unloaded. Two sections of pipe are available for use and each section is to be equipped with a pump located at the beginning of the section. Pertinent data for the piping system containing smooth pipe is as follows: Pipe 1 is 3-km long, has a diameter of 80 cm and its pump has an output of 1000 kPa. Pipe 2 is 11-km long, has a diameter of 100 cm and its pump has an output of 860 kPa (Figure 17.11). Section 2 is equipped with a heater that will heat the oil as needed to modify its flow properties. The oil leaves the tanker and enters the system at 20◦ C. The cargo space in the tanker is 160,000m3 . 3 km

8 km

100 cm 80 cm Section 1 FIGURE 17.11 Piping configuration for Design Example 11.

Section 2

Flow in Pipes and Pipe Networks

547 C = 1779 + 1.660 T°C J/kg-K

ρ

840

μ

6.0

830

5.0

820

ρ

4.0

Density, kg/m3

μ × 103, kg/m-s

7.0

810

μ 3.0

0

10

20

30 40 Temperature, °C

50

60

FIGURE 17.12 Thermal properties of 34◦ mid-continent crude oil. (Taken from Kern, D. Q., Process Heat Transfer, McGraw-Hill Book Company, New York, NY, 1951.)

The oil is a mid-continent crude (34◦ API) and thermal properties are provided in Figure 17.12. Determine (a) the capacity of the piping system, (b) the required power, if any, for heating the oil, and (c) the time required to completely empty the tanker using the pump at the entrance to Section 1.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) Data for the piping system is specified. (3) Pertinent properties of mid-continent crude are given in Figure 17.12. (4) Physical properties of the oil are uniform at any cross section. (5) Both pipes may be considered as smooth. Strategy Each pump is to be operated at its maximum capacity. We will use section 1 and section 2 in that order because the pump 1 operates at a higher pressure and section 1 is not equipped with a heater. One procedure is 1. Assume values of the oil velocity in section 1 and find, by trial and error, the velocity that will yield a head loss in the system of 1000 kPa.

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Introduction to Thermal and Fluid Engineering

2. Once the velocity is established, we will be able to determine the capacity of the system in both kilogram per second and gallons per minute. 3. We will see at 20◦ C whether the flow established in section 1 can be accommodated in section 2. If so, we will determine whether the head loss is section 2 is equal to the pump output of 875 kPa. If not, we will alter the temperature of the oil in section 2. This will adjust the flow so as to make the system optimum. In this case, “optimum,” means that both pumps are operating at their rated output pressures. Section 1 using properties from Figure 17.12 L = 3.0 km

d = 0.80 m

T = 20◦ C

L/d = 3000 m/0.80 m = 3750

 = 846.3 kg/m

3

 = 0.0062 kg/m-s

With the head lost given by Equation 17.6 hL = 2 f f

ˆ2 LV d g

values of Vˆ can be assumed so that hL = 2 f f

ˆ2 LV 2(3000 m) = f f Vˆ 2 = 764.53 f f Vˆ 2 d g (0.80 m)(9.81 m/s2 )

Then with Red =

d ˆ (846.3 kg/m3 )(0.80 m) ˆ V= V = 109, 200 Vˆ  0.0062 kg/m-s

where Vˆ is in m/s. Then, by trial-and error, thus with ˆ m/s Assume V,

8.0

7.25

7.00

Find Red × 103

130

87.3

79.2

77.0

Obtain, f f

0.00285

0.00300

0.00350

0.00318

h L , m of oil

313.76

146.8

126.6

120.8

2605

1218.7

1050.9

1003

h l , kPa

A=

7.05

 2  d = (0.80 m) 2 = 0.5027 m2 4 4

we have with Vˆ = 7.05 from the foregoing table m ˙ = AVˆ = (846.3 kg/m3 )(0.5027 m2 )(7.05 m/s) = 3000 kg/s (a) The capacity of the system is ˙ = Vˆ A = (7.05 m/s)(0.5027 m2 ) = 3.584 m3 /s ⇐ V Section 2, again with properties taken from Figure 7.12 L = 8.0 km m ˙ = 3000 kg/s

d = 1.00 m L/d = 8000 m/1.00 m = 8000

Flow in Pipes and Pipe Networks

549

With A=

 2  d = (1.00 m) 2 = 0.7854 m2 4 4

we have Assume T   Vˆ = m/A ˙

◦C

kg/m3 kg/m-s

Red × 106 ff hL

m/s

N/m2

20.0 870 0.0068

30.0 865 0.00525

40.0 860 0.0044

50 855 0.0038

4.577 0.5856 0.00303 893.5

4.603 0.7584 0.00292 856.3

4.630 0.9050 0.00286 843.6

4.657 1.048 0.00280 830.5

These data are plotted in Figure 17.13 where it may be observed that to meet the pressure at the outlet of pump 2, the temperature should be T = 37.5◦ C (b) With the specific heat taken from Figure 17.11 at 35◦ C is 1745 kJ/kg-K so that the heat supplied to the oil will be ˙ = mcT Q ˙ = (3172 kg/s)(1745 kJ/kg-K)(17.5◦ C) = 49.54 kW ⇐ (c) The time to empty the tanker at the volumetric flow rate established in (a) will be t=

160, 000m3 = 12.37 h ⇐ (3.584 m3 /s)(3600 s/h)

890 880 875 kPa Head Lost, kPa

870 860

34°C

850 840 830 820 810

10

20

30 40 Temperature, °C

FIGURE 17.13 Head loss as a function of temperature for pipe 2 in Design Example 11.

50

550

17.10

Introduction to Thermal and Fluid Engineering

Multiple-Path Pipe Systems

Many industrial piping systems involve multiple paths for fluid flow between source and destination. Such systems may be quite involved and any solution will likely be very laborious and time consuming. The design of new multiple-path systems or the analysis of the performance of an existing piping network will likely entail the use of a computer code. In this section, we will illustrate the process of analysis by using a relatively simple example. The steps in such an analysis are those that are used in more complex systems. A flow system, such as the one shown in Figure 17.14, is directly analogous to a parallel electrical circuit. There are two branches (numbered 1 and 2) between two nodes (A and B) for the multiple-path portion of the flow system shown. For this two-node system, it is clear that, for each branch, there is a common pressure difference, PA − PB , and a common elevation change, z B − z A. Furthermore, continuity at each node will require that ˙A = V ˙B = V ˙1 + V ˙2 V For each of the two branches, an equation of the form g(z B − z A) +

Vˆ 2B − Vˆ 2A PB − PA + + uB − u A = 0 2 

will apply and, as usual, the internal energy increase can be written as u B − u A = gh L = 2 f f

L ˆ 2  Vˆ 2 V + K d 2

Analysis of a two-node piping system will be of two general types: 1. Given the flow conditions at A, along with PA, the pipe size, geometry, and roughness, find PB . 2. Given PA and PB , along with the pipe size, geometry, and roughness, evaluate the ˙A + V ˙ B as well as V ˙ in each branch. total flow, V Example 17.4 presents the steps in analyzing a two-branch, two-node system. Branch 1

PB = ? A

PA = 965 kPa

Branch 2

FIGURE 17.14 A multiple-path piping system with two branches and two nodes.

B

Flow in Pipes and Pipe Networks

551

Example 17.4 In the piping system shown in Figure 17.14, the pressure at point A is 965 kPa

(gage) and the total flow rate of water at 16◦ C is 0.70 m3 /s. Pipe lengths and diameters are as follows: for pipe 1, L 1 = 550 m and d1 = 0.36 m and for pipe 2, L 2 = 850 m and d2 = 0.50 m. Neglect minor losses and determine the pressure at node B.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The flow is incompressible. (3) Minor losses are neglected. (4) Heat transfer is negligible. (5) Uniform properties exist at each cross section considered. The constraints on this system include continuity (i.e., the total flow in branches 1 and 2 must equal 0.70 m3 /s); the energy equation, which relates potential energy, kinetic energy, pressure change, and head loss along each of the branches connecting nodes A and B; and the frictional head loss, which must be compatible with Figure 17.4 for both branches. As a first step we can write the first law expression for each branch. For branch 1 g(z B − z A) +

Vˆ 21B − Vˆ 21A PB − PA + + u1B − u1A = 0 2 

(a)

g(z B − z A) +

Vˆ 22B − Vˆ 22A PB − PA + + u2B − u2A = 0 2 

(b)

and for branch 2

Noting that Vˆ 21B − Vˆ 21A Vˆ 2 − Vˆ 22A = 2B =0 2 2 and that the potential energy and flow energy change are the same for both branches, it is clear that u1 = u2 or h L1 = h L2

(c)

With the system constraints in mind, we can proceed with a trial-and-error solution to our problem. The procedure to be used is the following: ˙ 1 , through one branch and determine the associated head 1. Assume a flow rate, V loss. 2. Since, according to Equation (c), h L1 = h L2 , solve for the flow rate through the other ˙2. branch, V ˙ 1 /V ˙ 2 , adjust the values 3. Assuming that the actual flow divides in the same ratio as V ˙ ˙ ˙ ˙ ˙ for V1 and V2 so that V1 + V2 = VA. ˙ 1 and V ˙ 2 , solve for h L1 and h L2 . 4. With these values for V 5. If h L1 = h L2 , the solution is complete.

˙ 1 and return to step 2. 6. If h L1 =  h L2 , adjust the assumed value for V We now follow this procedure for the velocities specified in the problem statement.

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Introduction to Thermal and Fluid Engineering

˙ 1 = 0.20 m3 /s and solve for h L1 . With 1. Assume V ˙ 4V Vˆ = d 2 we have gh L1

L1 = 2 f f1 d1

 ˙ 2 4V1 32 L1 ˙ 2 = 2 f f1 5 V 1 2  d1 d1

Because the pipe is hydraulically smooth, we must determine f f 1 from Figure 17.4. The Reynolds number is evaluated as Re1 =

d1 Vˆ 1 d1 =  

 ˙  ˙1 4V1 4V = 2 d1  d1

For water at 16◦ C,  = 1.133 × 10−6 m/s2 , we have Re1 =

4(0.20 m3 /s) = 6.25 × 105 (0.36 m)(1.133 × 10−6 m2 s)

The corresponding value of f f 1 from Figure 17.4 is f f 1 = 0.0031 The value of gh L1 can now be evaluated: gh L1 =

32 (550 m) (0.0031) (0.020m3 /s) 2 = 36.57 m2 /s2 2  (0.36 m) 5

˙ 2 from the expression 2. With gh L1 = gh L2 = 36.57 m2 /s2 , we can calculate V gh L2 =

32 L2 ˙ 2 f f2 5 V 2 2 d2

˙ 2 , we have Solving for V 2 2 2 5 −4 ˙ 22 = (36.57 m /s ) (0.50 m) = 4.147 × 10 V 32(850 m) f f 2 f f2

By trial and error, this can be solved to provide ˙ 2 = 0.372 m3 /s V We note that ˙1 + V ˙ 2 = 0.20 m/s + 0.372 m/s = 0.572 m3 /s V 3. Adjusted values are determined as   0.20 ˙ V1 = (0.70 m3 /s) = 0.245 m3 /s 0.572 and ˙2 = V



 0.372 (0.70 m3 /s) = 0.455 m3 /s 0.572

Flow in Pipes and Pipe Networks

553

4. The corresponding head loss values are determined once more Re1 =

4(0.245 m3 /s) = 7.644 × 105 (0.36 m)(1.133 × 10−6 m2 /s)

with the corresponding value of f f 1 from Figure 17.4 of f f 1 = 0.00305 The value of gh L1 is now evaluated as 32 (550 m) (0.00305) (0.245 m3 /s) 2 = 54.0 m2 /s2 2 (0.36 m) 5

gh L1 =

These same steps carried through for pipe 2 yield Re2 = 8.356 × 105 and f f 2 = 0.0030 so that gh L2 =

32 (850 m) (0.0030) (0.455 m3 /s) 2 = 54.7 m2 /s2 2  (0.50 m) 5

5. These two values of head loss compare within 1.3%, which is an acceptable accuracy. ˙ 1 and V ˙ 2 obtained in step 3 are satisfactory. We thus conclude that the values for V 6. With the average value of gh L taken as 54.35 m2 /s2 , we may use either Equation (a) or (b) to evaluate PB . The solution is PA − PB = g(z B − z A) + u B − u A  = (9.81 m/s2 )(12.2 m) + 54.35 m2 s2 = 174.0 m2 /s2 The final result is now obtained as PB = PA − (174 m2 /s2 )

pjwstk|402064|1435600795

= 965 kPa − (1000 kg/m3 )(174 m2 /s2 ) = 791 kPa ⇐ The foregoing example has illustrated a solution procedure for a representative twobranch pipe flow problem of the type where the flow conditions at A, along with PA, the pipe size, geometry, and roughness are given and PB is to be found. The procedure for solving a problem where PA and PB , along with pipe size, geometry, and roughness are ˙ in each branch is to be found, would follow a similar given and the total flow as well as V thought process. As stated earlier, the two-node, two-branch problem is as simple as multiple-path analysis gets. Myriad paths and nodes may exist in a real system. The solution for such a flow network is applied to analyze the individual relationships that compound to become a complex network.

554

17.11

Introduction to Thermal and Fluid Engineering

Summary

This chapter has dealt with the common and very important subject of pipe flow. Initially, the relationships for frictional pressure drop associated with both laminar and turbulent flow were examined. It was determined that the Fanning friction factor is functionally related to the Reynolds number, Red , and the relative roughness, e/d, as represented by the friction factor plot of Figure 17.4. A means for accounting for minor losses associated with valves, fittings, and entrance and exit effects was also considered. A suitable relationship, Equation 17.19, along with coefficients listed in chapter tables and plots, is available to compute these effects. The chapter concluded with a series of problems dealing with single- and multiplepath flows. Most problems of these type involve trial and error to match the constraints of friction factor along with the satisfaction of mass conservation and the first law of thermodynamics.

17.12

Problems

Mass Conservation Revisited 17.1: A circular pipe increases from an inlet (section 1) with a diameter of 30 cm to an outlet (section 2) with a diameter of 45 cm. Water flows through the pipe at a velocity of 5 m/s at section 2. Determine (a) the velocity at section 1, (b) the volumetric flow rate at section 1, (c) the volumetric flow rate at section 2, and (d) the mass flow rate. 17.2: A gas flows through a tapering square conduit. At section 1, the conduit has sides that are 0.08 m in length and at section 2, the sides are 0.24-m long. The gas velocity is 8.25 m/s at section 1 and its density is 1.092 kg/m3 . At section 2, the velocity is 1.98 m/s. Determine the mass flow rate of the gas and its density at section 2. 17.3: Water enters a tank at sections 1 and 2 and leaves at section 3. The velocity of the water entering the tank at section 1 is 4.875 m/s and the volumetric flow rate at ˙ = 0.0125 m3 /s. The diameters of the conduits at sections 1 and 3 are section 2 is V 4 cm and 6 cm, respectively. The water level in the tank remains constant. Determine the exit velocity, Vˆ 3 . 17.4: Air at 27◦ C and 108 kPa flows at 1.5625 kg/s through a circular annular passage with an inner diameter of 16 cm and an outer diameter of 32 cm. Determine (a) the average velocity and (b) the volumetric flow rate. 17.5: Oil with a specific gravity of 0.858 flows through a 75-cm-diameter pipe at a flow rate of 8760 gal/min. Determine (a) the volumetric flow rate in m3 /s, (b) the velocity, and (c) the mass flow rate. 17.6: A mixing chamber has an inlet (section 1) and an outlet (section 2). A third conduit, which can be used as either an inlet or an outlet (section 3), is connected to the chamber. At section 1, the diameter is 5 cm and the volumetric flow rate is 0.60 m3 /s. At section 2, the diameter is 7.5 cm and the velocity of egress is 12 m/s. If the diameter at section 3 is 2.5 cm, determine whether the flow at section 3 is into or out of the mixing chamber. 17.7: Consider a plunger-exit pipe arrangement in which the plunger at section 1 is being pushed into a tank with a fluid having a specific gravity of 0.72. At section 1 the

Flow in Pipes and Pipe Networks

555

velocity of the plunger is 6.25 cm/s. A pipe with a diameter of 2 cm leads from the tank at section 2. Determine the mass rate of flow at section 2. 17.8: A gasoline pump is used to fill a 100-L tank in 90 s. The pump has an exit diameter of 3.75 cm. Determine the exit velocity from the pump. Flow Regimes 17.9: Unused engine oil at 27◦ C flows in a 24-cm pipe. Determine the maximum velocity to assure laminar flow. 17.10: For a Reynolds number of 300,000, determine the velocity of an airstream at 27◦ C flowing past a 12-cm-diameter sphere. 17.11: Air flows in a circular pipe at a rate of 0.04 kg/s; the air temperature is 27◦ C. Determine the minimum allowable pipe diameter for the flow to remain laminar. 17.12: Water, at 15◦ C, flows in an annular space between two concentric pipes. The inner and outer diameters of the annular region are 7.5 cm and 11.5 cm, respectively. (a) Determine the volumetric flow rate, in m3 /s, that will be achieved for a water velocity of 3 m/s and (b) specify whether this flow is laminar or turbulent. 17.13: For 15◦ C water flowing in the annular region specified in Problem 17.12, determine what maximum water velocity can be obtained if the flow is to be laminar. 17.14: Solve Problem 17.13 if the fluid is air at 25◦ C. 17.15: Rework Problem 17.10 for flowing fluids of (a) water at 20◦ C and (b) nitrogen at 27◦ C. 17.16: For fluid flow at 900 cm3 /s through a 6.5-cm-diameter duct, determine whether the flow is laminar or turbulent for each of the following fluids at 20◦ C: (a) air, (b) carbon dioxide, (c) ethylene glycol, (d) water, (e) unused engine oil, and (f) glycerin. 17.17: SAE 50 oil flows through a 10-cm-diameter pipe. Determine the flow rate, in m3 /s, for transition to turbulent flow for (a) 27◦ C and (b) 100◦ C. 17.18: For glycerin flowing through a 5-cm-diameter pipe, determine the volumetric flow rate, in m3 /s, for transition to turbulence at (a) 27◦ C and (b) 47◦ C. 17.19: The Reynolds number for a fluid in a 25-cm-diameter pipe is 2000. A pipe that is 15-cm diameter forms an extension to the 25-cm-diameter pipe. The flow is incompressible. Determine the Reynolds number in the 15-cm-diameter pipe. 17.20: Water flows through two pipes 1 and 2 that intersect to form a single conduit (pipe 3). Pipe 1 has a diameter of 5 mm and the volumetric flow rate in the two pipes are equal. The diameters of pipes 2 and 3 are 4 mm and 6 mm, respectively. The water ˙ 2 , in pipe 2 in temperature is 27◦ C. Determine the maximum volumetric flow rate, V order to keep the flow in pipe 3 laminar. 17.21: Compare the maximum velocity to yield laminar flow in a 15-cm pipe for (a) a fuel oil at 15◦ C with a kinematic viscosity of  = 4.42 × 10−6 m/s2 and (b) water at 15◦ C. 17.22: For laminar flow conditions, determine the diameter of the pipe needed to deliver 10 gal/min of a fuel oil having a viscosity of  = 5.92 × 10−7 m/s2 . 17.23: A fluid with a kinematic viscosity of  = 1.35 × 10−5 m/s2 flows through a 24-cmdiameter pipe. Determine the maximum velocity to keep the flow laminar. 17.24: Determine the Reynolds number of oil flowing through a 42-cm-diameter pipe if the oil has a specific gravity of 0.852, a dynamic viscosity of 0.0258 kg/m-s and a volumetric flow rate of 0.385 m3 /s.

556

Introduction to Thermal and Fluid Engineering

17.25: Determine the type of flow occurring in a 30-cm-diameter pipe when (a) water at 15◦ C flows at a velocity of 1.25 m/s and (b) a fuel oil with  = 2 × 10−5 m/s2 flowing at the same velocity. 17.26: An oil with a specific gravity of 0.862 and a kinematic viscosity of  = 2 × 10−5 m/s2 flows through a pipe that has a diameter of 25 cm. The volumetric flow rate is 0.625 L/s. Determine whether the flow is laminar or turbulent. Head Loss—Laminar Flow 17.27: An oil with a density of 910 kg/m3 and a kinematic viscosity of 7.4×10−6 m/s2 flows through a horizontal tube 6 mm in diameter at a volumetric flow rate of 0.0838 m3 /h. Determine the pressure drop that will be achieved if the tube is 10-m long. 17.28: For the 10-m-long tube described in Problem 17.27, determine the maximum volumetric flow rate in m3 /s to keep the flow laminar. 17.29: A tube used in lubrication is 75-cm long and has a diameter of 2.5 mm; the oil properties are as given in Problem 17.27. Determine the flow rate of the oil when the pressure drop is 100 kPa. 17.30: A viscometer for measuring the viscosity of liquids commonly consists of a relatively large reservoir connected to a relatively small tube as shown in Figure P17.30. An oil flows out of the viscometer at a rate of 29 cm3 /s into a 1.8-mm-diameter tube. Determine the viscosity of the oil.

Oil

8 cm

55 cm

FIGURE P17.30

17.31: For the viscometer specified in Problem 17.30 and sketched in Figure P17.30, water at 15◦ C is used as the viscous fluid. Determine the expected flow rate of the water in cm3 /s. 17.32: Oil with a kinematic viscosity of 4.5 × 10−6 m/s2 and a density of 800 kg/m3 is to be pumped at a rate of 3.5 m3 /s through a 260-km-long pipeline, 62 cm in diameter, made of commercial steel. Determine the pumping power required to achieve these conditions. 17.33: Rework Problem 17.32 when the flowing fluid is water at 40◦ C. 17.34: A 65-km-long pipe line delivers petroleum at a rate of 5000 barrels per day (795 m3 /day) with a pressure drop, between origin and destination, of 4.07 MPa. Assuming that the flow in the pipe to be at its maximum value for laminar flow, determine the diameter of the pipe. 17.35: For the case in Problem 17.34, consider that a parallel section of the same diameter pipe is added over the last 40 km of the system and, still assuming the flow in the system to be laminar, and the pressure drop for the total system to remain at 4.07 MPa, determine the new capacity of the piping system.

Flow in Pipes and Pipe Networks

557

17.36: Solve Problem 17.34 for the case with two additional sections of the same diameter pipe added over the last 40 km of the pipeline. All other conditions remain as given in Problem 17.34. Determine the new capacity of the piping system. 17.37: A liquid flowing under the influence of gravity down an inclined plane surface, with its surface exposed to the atmosphere, is referred to as a falling film. For such a film in steady, laminar, and incompressible flow, the velocity profile between the wall and the free surface varies with the distance, y, as shown in Figure P17.37, according to the expression gL 3 sin  Vˆ x = 



y y2 − L 2L 2



Determine the values of (a) Vˆ x,max and (b) Vˆ x,avg . y Free Surface x

V x(y) L

θ

FIGURE P17.37

17.38: A horizontal capillary tube with an inside diameter of 5 mm is used to measure the viscosity of thick fluids such as oils. In such a system, the pressure gradient is measured to be 380 kPa/m of tube length when the flow rate of the liquid is 0.073 m3 /h. Determine the fluid viscosity. 17.39: The capillary tube viscometer of Problem 17.38 is to be used with glycerin at 20◦ C, determine the flow rate, in m3 /h, that is achieved for a Reynolds number of 2000. 17.40: Water is being siphoned from a reservoir as shown in Figure P17.40. For h = 100 cm, will the flow be laminar or turbulent?

h

Flow FIGURE P17.40

17.41: An oil, having a density of  = 870 kg/m3 and a dynamic viscosity of  = 0.065 kg/m-s, is pumped through a horizontal pipe, 1.2-km long, with an input power of 1 kW. If laminar flow is to be maintained, determine (a) the maximum flow that can be achieved and (b) the corresponding pipe diameter.

558

Introduction to Thermal and Fluid Engineering

17.42: The piston shown in Figure P17.42 provides constant flow through the hypodermic needle. Determine the force necessary to achieve a flow rate of 0.15 cm3 /s, with a fluid having a density  of 870 kg/m3 and a viscosity of  = 0.002 kg/m-s. d2 = 1 cm d1 = 0.25 mm

1.5 cm

F

3 cm FIGURE P17.42

Head Loss—Other Types of Flow 17.43: Water at 20◦ C flows through a cast-iron pipe at a velocity of 3.2 m/s. The pipe is 400 m long and has a diameter of 15 cm. Determine the head loss due to friction. 17.44: A 2.44-m-diameter pipe carries water at 15◦ C. The head loss due to friction is 0.500 m per 300 m of pipe. Determine the volumetric flow rate of the water leaving the pipe. 17.45: Water at 20◦ C is being drained from an open tank through a cast-iron pipe 60 cm in diameter and 40-m long. The surface of the water in the pipe is at atmospheric pressure and at an elevation of 45.9 m, and the pipe discharges to the atmosphere at an elevation of 30 m. Neglecting minor losses due to configuration, bends, valves, and fittings, determine the volumetric flow rate of the water leaving the pipe. 17.46: Unused engine oil is being pumped at a volumetric flow rate of 0.20 m3 /s through a level, 15-cm-diameter wrought-iron pipe. Determine the pressure loss in Pa per diameter of pipe. 17.47: SAE 50 oil at 20◦ C is to be pumped at a volumetric flow rate of 0.275 m3 /s through a level cast-iron pipe. The allowable pipe friction loss is 70,000 Pa per 300 m of pipe. Determine the size of commercial pipe to be used. 17.48: Water at 20◦ C is to be pumped through a welded-steel pipe (e = 0, 457 × 10−4 m) at 4.25 m/s. The pipe is 400-m long and has a diameter of 17.5 cm. Determine the head loss due to friction. 17.49: Gasoline with a specific gravity of 0.714 and a dynamic viscosity of 2.96×10−4 kg/m-s is being discharged from an inclined pipe in which section 1, at the inlet, is 16-m higher than the outlet. The pressure at section 1 is 2400 kPa and the discharge is at atmospheric pressure. The pipe roughness, e, is 0.480 mm and the pipe is 950-km long. Determine the diameter needed to yield a volumetric flow rate of 0.125 m3 /s. 17.50: Water at 20◦ C flows through a 10-cm-diameter cast-iron pipe at a velocity of 4.8 m/s. Determine (a) the pressure loss per 100 m of pipe and (b) the power lost due to friction. 17.51: A 15-cm wrought-iron pipe is to carry water at 20◦ C. Assuming level pipe, determine the volumetric flow rate at the discharge if the pressure loss is not to exceed 32.5 kPa per 100 m of pipe length. 17.52: An oil having a kinematic viscosity of 8.30 × 10−5 m/s2 flows through a 600-m horizontal pipe with a diameter of 7.5 cm. For an upstream pressure of 33.5 kPa and a downstream pressure of 24.0 kPa, determine the mass flow rate in kg/s. 17.53: Alcohol with a specific gravity of 0.605 and a kinematic viscosity of 5 × 10−7 m/s2 is drawn from a tank through a hose having an inside diameter of 2.5 cm as shown

Flow in Pipes and Pipe Networks

559

in Figure P17.53. The hose has a relative roughness of e = 0.0004 m. The total length of the hose is 8.75 m and we are to neglect minor losses due to configuration, bends, valves, and fittings. Determine (a) the volumetric flow rate of the alcohol leaving the hose and (b) the minimum pressure in the hose.

3m

Alcohol SG = 0.605

2.5 cm

8.75 m ν = 5 × 10–7 m2/s FIGURE P17.53

17.54: A level 100-m-long water pipe has a manometer at both the inlet and the outlet. The manometers indicate pressure heads of 1.5 and 0.2 m, respectively. The pipe diameter is 15 cm and the pipe roughness is 0.0004 m. Determine the mass flow rate in the pipe in kg/s. 17.55: Level wrought-iron pipe having a diameter of 4.5 cm and a length of 2 km carries water at 20◦ C at a velocity of 7 m/s. Determine (a) the head loss and (b) the pressure drop. 17.56: Mercury (SG = 13.6,  = 1.56×10−6 kg/m-s) flows through 2.75 m of 6-mm-diameter glass tubing with a velocity of 2 m/s. Determine (a) the head loss and (b) the pressure drop. 17.57: Acetic acid (SG = 0.86,  = 2.60 × 10−6 m/s2 ) flows at 0.325 m/s through cast-iron pipe having a diameter of 15 cm. The pipe is 0.626-m long and slopes upward from inlet to outlet at an angle of 12◦ between inlet and outlet. Determine (a) the head loss and (b) the pressure change between inlet and outlet. 17.58: Determine the depth of water behind the dam in Figure P17.58 that will provide a flow rate of 5.675 × 10−4 m3 /s through an 18-m long, 1.25-cm-diameter commercial steel pipe.

h 1.25 cm . V = 5.675 × 10–4 m3/s 18 m FIGURE 17P.58

17.59: Using minor loss coefficients for the elbows, globe valve and entrance of 0.90 (each), 10.0, and 0.50, respectively, determine the volumetric flow rate in the 1.5-cm pipeline in Figure P17.59 for h = 12 m as shown, the commercial steel pipe is 140-m long. 12 m

Water at 27°C

25 m

75 m 40 m

FIGURE P17.59

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Introduction to Thermal and Fluid Engineering

17.60: A pipeline consists of 75 m of 10-cm-diameter steel pipe. The pipeline includes a 90◦ bend, a fully open gate valve, an additional 35 m of 10-cm pipe with a 20◦ taper, an abrupt contraction to 7.5-cm, and an additional 20 m of 7.5-cm steel pipe. The discharge rate through this piping system is 0.0425 m3 /s. Determine the head lost. 17.61: Water flows at 20◦ C from reservoir 1 to reservoir 2 through 150 m of 10-cm diameter of welded-steel pipe as indicated in Figure P17.61. Determine the volumetric flow rate in the pipe. 1 2m

Water at 20°C

10 cm 2

8m 150 m

2m

FIGURE P17.61

17.62: Figure P17.62 shows two reservoirs holding water at 40◦ C. They are connected by 35 m of commercial steel pipe of diameter, 10 cm, containing a sharp-edged entrance, two 90◦ elbows and a gate valve that is 50% closed. Determine the volumetric flow rate in the piping system between the reservoirs when the water surface elevation in the reservoirs are as shown.

8m

Total Pipe Length = 35 m

Water at 40°C

8m

6m

10 cm

FIGURE P17.62

17.63: A piping system carries water from a reservoir and discharges it as a free jet as indicated in Figure P17.63. Determine the flow through the 20-cm-commercial steel pipe with the fittings shown.

40 m

Water at 20°C

20 m

28.28 m

10 m 45° 20 cm

20 m FIGURE P17.63

20 m

Flow in Pipes and Pipe Networks

561

17.64: The piping system shown in Figure P17.64 is to be used to transport oil from tank 1 to tank 2 at a maximum volumetric flow rate of 0.0285 m3 /s. Determine the diameter of the steel pipe required.

SAE 50 Oil 17°C

6m

8m

1

6m

12 m

2

FIGURE P17.64

Multiple Pipe Systems 17.65: Water flows at a volumetric flow rate of 0.225 m3 /s and 20◦ C from reservoir 1 to reservoir 2 through three concrete pipes connected in a series. Pipe 1 is 1000-m long and has a diameter of 16 cm. Pipe 2 has a length of 1500 m and a diameter of 20 cm. Pipe 3 is 800-m long and the diameter is 18 cm. Neglecting minor losses, determine the difference in surface elevations. 17.66: A piping system consists of three pipes in a series. The total pressure drop is 160,000 Pa and the difference in elevation is 5 m. Data for the three pipes are Pipe

Length, m

Diameter, cm

Roughness, mm

1 2 3

125 150 100

8 6 4

0.240 0.120 0.200

For water at 20◦ C, neglect minor losses and determine the volumetric flow rate in the system. 17.67: Two concrete pipes are connected in series. The flow rate of water at 20◦ C through the pipes is 0.15 m3 /s, with a total head loss of 15 m for both pipes. Each pipe has a length of 312.5 m and a relative roughness of 0.0035 m. Neglecting minor losses, if one pipe has a diameter of 30 cm, determine the diameter of the other. 17.68: A piping system containing a loop is shown in Figure P17.68. The total volumetric flow of water at 20◦ C is 0.20 m3 /s. Determine the volumetric flow rate in pipes C and D and the head lost from point 1 to point 2. Cast-Iron Pipes

Pipe – C d = 15 cm L = 800 m

Water @ 20°C . V = 0.20 m3/s

2

1

Pipe – D d = 12 cm L = 1200 m FIGURE P17.68

562

Introduction to Thermal and Fluid Engineering

17.69: A commercial pipe network contains two branches in parallel as shown in Figure P17.69. A volumetric flow of 0.575 m3 /s of a liquid with  = 1000 kg/m3 and  = 0.1025 × 10−4 m/s2 passes through the parallel combination. The pressure is 700 kPa at node 1. Neglecting minor losses, determine the pressure at node 2. Commercial Steel Pipes

d = 12 cm L = 1000 m

. V = 0.575 m3/s P2 = ?

P1 = 700 kPa

d = 18 cm L = 1000 m FIGURE P17.69

17.70: A piping system consists of three pipes in parallel with a total loss of head of 21 m. Data for the three pipes are as follows: Pipe

Length, m

Diameter, cm

Roughness, mm

1 2 3

100 150 80

8 6 4

0.240 0.120 0.200

For water at 20◦ C, neglect minor losses and determine the volumetric flow rate in the system.

18 Fluid Machinery

Chapter Objectives •

To describe the various types of fluid machines.

•

To provide the theoretical considerations for centrifugal pumps.

•

To discuss the problem of cavitation and the use of the net positive suction head.

To show how the performance of a system is matched to the performance of a centrifugal pump the system contains. • To derive the scaling laws for pumps and fans. •

•

18.1

To discuss axial and mixed flow pumps and turbines.

Introduction

The term fluid machinery is commonly used to categorize mechanical devices that exchange fluid energy and mechanical work. When mechanical energy is applied to a fluid, producing flow or higher pressure—or both, the machine is a pump. When the reverse is true, and fluid energy drives the machine to produce mechanical work, we call the machine a turbine. There are two principal types of fluid machines, positive displacement machines and turbo machines. In a positive displacement machine, a fluid is confined in a chamber whose volume is varied. The human heart and a bicycle tire pump are both positive displacement pumps. Examples of such devices are shown in Figure 18.1. Turbo machines involve rotary motion as the name implies. Window fans and aircraft propellers are examples of unshrouded turbo machines. Pumps used with liquids generally have a shroud that surrounds the impeller and thus contains and guides the flow. Two categories of pumps are shown in Figure 18.2, the radial flow pump and the axial flow pump. The designations radial and axial refer to the direction of flow relative to the axis of rotation of the blades. The term pump is generally used with flows of liquids. When a gas or vapor is the fluid of interest, the following terms apply: Fans, which are generally associated with relatively low pressure changes with P in the neighborhood of 35-cm H2 O (1/2 psi). • Blowers, which are in positive and variable displacement configurations with P up to 2.8-m H2 O (40 psi). •

563

564

Introduction to Thermal and Fluid Engineering Aorta

To Body

Pulmonary Artery

Right Atrium

To Lungs Oxygenated Blood from Lungs

Left Atrium Venous Blood

Suction

Right Ventricle

Discharge

Left Ventricle

FIGURE 18.1 Examples of positive displacement pump configurations. •

18.2

Compressors, which are in positive and variable displacement configurations with delivery pressures as high as 69 MPa (104 psi).

The Centrifugal Pump

18.2.1 Introduction We will first discuss the operation of centrifugal pumps that are commonly used in domestic and industrial applications. A performance curve for a centrifugal pump is basically a plot of pressure rise (or head increase) between pump inlet and discharge as a function of flow rate. Typical performance for a centrifugal pump is shown in Figure 18.3. For comparison purposes, the performance of a positive displacement pump is also shown. In the United States, conventional units for head are feet of water and for flow rate, gallons per minute. In keeping with our general use of SI units in this text, we will use meters of water (m H2 O) and m3 /s, respectively, in all subsequent discussion.

Fluid Machinery

565

Outlet Rotor Inlet

Housing or Casing (a) Radial Flow Fan Rotor

Inlet

Outlet Housing or Casing

Stator (b) Axial Flow Fan

Pressure Rise or Head Increase

FIGURE 18.2 (a) Radial flow and (b) axial flow configurations.

0

Positive Displacement Pump Low μ

Low μ

High μ

Centrifugal Pump High μ

Discharge

FIGURE 18.3 Performance of a typical centrifugal pump. The symbol, , indicates viscosity.

566

Introduction to Thermal and Fluid Engineering

The shape of a performance curve, and values for pressure head and flow rate, are unique for a given pump configuration at a particular speed of rotation. Considerable art and empirical judgment are used in designing pumps. Once a pump has been fabricated and a baseline performance curve has been determined, rules exist for predicting performance at other rotational speeds and characteristic sizes and for other fluids. We will address these pump scaling laws in Section 18.5. 18.2.2 Theoretical Considerations Figure 18.4 represents the impeller blade configuration for backward curved blades. The impeller rotates, as indicated, with an angular velocity, . Our analysis will involve the moment-of-momentum expression presented in Chapter 14. The governing equation is 

M=



(r × V e ) m ˙e −

e

 d (r × V i ) m ˙ i + (r × V cv )mcv dt i

(15.5)

In the present analysis we will consider the coordinate origin to be at the axis of rotation so that rotation is about the z axis. For one-dimensional steady flow, the scalar form of Equation 15.5, written for the z-direction, is        i )  m Mz = (r × V e )  m ˙e − (r × V (18.1)  ˙i z

e

i

z

Equation 18.1 will be solved for the control volume shown with the dashed lines in Figure 18.4 (left). Note that the entering flow is along the axis of rotation (the z direction) and that the exiting flow occurs along the outer circumference of the impeller. An expanded view of the velocity components entering and exiting the impeller is shown in Figure 18.5. Equation 18.1 can now be evaluated: by definition, Vˆ 1 = Vˆ z1 = 0   ⎤ ⎡  r θ z   r θ z      Mz = m ˙ ⎣ r 0 0  −  r 0 0  ⎦  Vˆ r Vˆ  Vˆ z   Vˆ r Vˆ  Vˆ z  2 1 z or ˙ 2 Vˆ 2 Mz = m[(r ˙ 2 Vˆ 2 ) − 0] = Vr

(18.2)

2

Casing 1 Impeller

Expanding Area Scroll Control Volume FIGURE 18.4 Flow through the impeller of a centrifugal pump.

Fluid Machinery

567 Vr2 = r2ω Vn2 Vb1 Vb2 β2

r2

β1

r1 FIGURE 18.5 Velocity diagrams for conditions at the exit of a centrifugal pump impeller.

The challenge here is in the evaluation of Vˆ 2 . By definition, Vˆ 2 is the tangential component of the exiting fluid stream relative to the fixed coordinate system. The velocity diagram in Figure 18.6 is useful in evaluating this quantity. The absolute velocity of the exiting flow, V 2 , is the vector sum of the velocity relative to the impeller blade and the blade tip velocity relative to our fixed coordinate frame. In terms of the dimensions shown in Figure 18.6, we express the following: The normal velocity of flow for blade length, L, at r2 Vˆ n2 =

˙ V 2r2 L

The velocity of flow along the blade Vˆ n2 Vˆ b2 = sin 2 The blade tip velocity Vˆ t2 = r2  Vt1 = r1ω Vr1

Vb1 β1

r1

FIGURE 18.6 Velocity relationships at blade inlet.

568

Introduction to Thermal and Fluid Engineering

We can now evaluate the quantity, Vˆ 2 , which is normal to r2 as Vˆ n2 Vˆ 2 = r2  − Vˆ b2 cos 2 = r2  − cos 2 sin 2 or Vˆ 2 = r2  − The moment, Mz , is thus,

˙ V cot 2 2r2 L

 ˙ Mz = Vr2 r2  −

˙ V cot 2 2r2 L

and the power delivered to the fluid is by definition, Mz  or  ˙ V ˙ ˙ W = Vr2  r2  − cot 2 2r2 L

(18.3)

(18.4)

(18.5)

We will keep this result in mind as we next examine our pump from an energy point of view. The first law of thermodynamics applied to the same control volume as before will yield the following. Beginning with Equation 13.11 d ˙ cv − W ˙ cv E cv = Q dt



   Vˆ i2 Vˆ 2e + m ˙ i hi + + gzi − m ˙ e he + + gze 2 2 e i

(13.11)

we have, for steady flow without heat transfer,

Vˆ 22 − Vˆ 21 ˙ cv = m −W ˙ h2 − h1 + + g(z2 − z1 ) 2 It is customary to neglect the relatively small differences in velocity and elevation between points 1 and 2, that is Vˆ 22 − Vˆ 21 ≈ 0

and

z2 − z1 ≈ 0

The expression that remains is   P2 − P1 ˙ cv = m −W ˙ u2 − u1 + 

(18.6)

We recall that the term u2 − u1 represents the loss in head due to friction and other irreversible effects. Thus, Equation 18.6 provides an expression for the net pressure head produced in the pump ˙ cv P2 − P1 W =− − hL  m ˙

(18.7)

Equations 18.5 and 18.7 can be combined to express the net pressure head as a function blade angle, flow rate, angular velocity, and impeller radius. Head losses must be determined experimentally.

Fluid Machinery

569 Shutoff Head

Head Brake Horsepower, bhp Efficiency, η

Head

Efficiency

Brake Horsepower

Normal or Design Flowrate 0

. Flowrate, V

0

FIGURE 18.7 Performance curves for a centrifugal pump impeller.

The foregoing results can now be used to evaluate another important performance parameter, the efficiency. In broad terms, efficiency is expressed as the ratio of actual output to required input. For the centrifugal pump, the efficiency, designated as , will be =

power added to the fluid shaft power to the impeller

The power required by the fluid has been expressed as   P2 − P1 ˙ fluid = m W ˙ 

(18.8)

so that the efficiency can be expressed as =

m( ˙ P2 − P1 ) ˙ cv W

(18.9)

The performance curves presented in Figure 18.4 can now be enhanced to include the efficiency and the input shaft power that is generally referred to as brake power. Figure 18.7 includes these quantities in addition to the developed head. For minimal friction loss at the radial location, r1 , where the inlet flow enters the impeller, it is desirable that flow relative to the blade be along the surface, oriented at the angle 1 . An expanded view of this section is shown in Figure 18.8. The design point for minimum losses is achieved when Vˆ b1 cos 1 = r1  or, equivalently, when sin 1 Vˆ r 1 = Vˆ b1 sin 1 = r1  cos 1 and, finally, when Vˆ r 1 = r1  tan 1

(18.10)

570

Introduction to Thermal and Fluid Engineering r1ω

Vr1 Vb1

ω

β1

r1

Blade

FIGURE 18.8 Expanded velocity diagram of conditions in Figure 18.6.

The following example demonstrates how the foregoing analysis relates to centrifugal pump performance.

Example 18.1 A centrifugal water pump with r2 = 9 cm, r1 = 5.125 cm, and blade length,

L = 4.5 cm, is designed with backward curved blades having 1 = 30◦ and 2 = 20◦ . For a rotational speed of 1400 rpm, and given  = 1000 kg/m3 for water, estimate (a) the designpoint volume flow rate, (b) the power gained by the water, and (c) the discharge pressure head.

Solution Assumptions and Specifications (1) Steady flow exists. (2) The flow is incompressible. (3) There is uniform flow at the inlet and outlet sections. (4) Torques due to body and surface torques may be neglected. (a) To establish the design-point volume flow rate, we note that for operation at the design condition, Equation 18.10 must be satisfied. Thus, we have  = (1400 rev/min)(1 m/60 s)(2 rad/rev) = 146.6 rad/s Vˆ r 1 = r1  tan 1 = (0.05125 m)(146.6 rad/s) tan 30◦ = 4.338 m/s The corresponding flow rate is ˙ = 2r1 L Vˆ r 1 V = 2(0.05125 m)(0.045 m)(4.338 m/s) = 0.0629 m3 /s

(996 gpm) ⇐

(b) The power added to the fluid can be determined using Equation 18.5  ˙ = Vr ˙ 2  r2  − W

˙ V cot 2 2r2 L

(18.5)

Fluid Machinery

571

With ˙ 2 = (1000 kg/m3 )(0.0629 m3 /s)(146.6 rad/s)(0.09 m) Vr = 830.29 kg-m/s2 and ˙ V 0.0629 m3 /s = = 2.47 m/s 2r2 L 2(0.09 m)(0.045 m) we obtain ˙ = (830.29 kg-m/s2 )[(0.09 m)(146.7 rad/s) − (2.47 m/s) cot 20◦ ] W = 5314 W ⇐ This is equivalent to 7.13 hp (c) The ideal discharge pressure (when losses are neglected) can be determined using Equation 18.7. The discharge pressure head is expressed as ˙ cv P2 − P1 W  g mg ˙ 

(5314 W)(1 kg-m2 /W-s3 ) (1000 kg/m3 )(0.0629 m3 /s)(9.81 m/s2 )

 8.61 mH2 O ⇐ This result is the maximum that can be achieved. The actual value will be less than this due to friction and other irreversible losses.

18.3

The Net Positive Suction Head

A major cause of degradation of pump performance is the presence of cavitation. Cavitation occurs when the liquid being pumped vaporizes or boils whereupon the vapor bubbles that form cause a decrease in efficiency and structural damage to the pump, which sometimes results in catastrophic failure. The parameter that characterizes the potential for cavitation to occur is the net positive suction head (NPSH). At the impeller, the total head on the suction side, which is the low pressure location at which cavitation will first occur, is related to the NPSH according to NPSH +

Vˆ 2 Pv Pi = i + g 2g g

(18.11)

where Vˆ i and Pi are the velocity and pressure at the pump inlet and Pv is the liquid vapor pressure. Values of NPSH are generally determined experimentally for a given pump over a range of flow rates. Typically, the variation of NPSH appears as shown in Figure 18.9. In the representative installation indicated in Figure 18.10, the liquid being pumped is drawn from a reservoir whose level is the distance, z, below the pump inlet. An energy

65

63

65

7

63

60 55

300

6

40 b 50 hp

200

30

15

25

100

20 NPSH R

0

15 10 5

0

40

80

120

160 200 240 Capacity, gal/min

280

320

0

NPSHR, ft

Head, ft

400

60

8 in. dia

500

55

Introduction to Thermal and Fluid Engineering 50%

572

FIGURE 18.9 Pump performance including NPSH behavior.

balance between the reservoir surface and the pump inlet yields Vˆ 2  Patm P2 = z2 + + 2+ hL g g 2g

(18.12)

 where h L represents the head losses between points 1 and 2. Equation 18.12 can be rewritten as  Vˆ 2 P2 Patm + 2 = − z2 − hL g 2g g and, with the incorporation of Equation 18.11 for the NPSH, we have NPSH =

Patm Pv  − z2 − − hL g g

(18.13)

From these results, and a performance plot such as Figure 18.9, a pump system designer can establish operating conditions to avoid the possibility of cavitation. The value of NPSH 2 z1

P1 = Patm 1

FIGURE 18.10 Pump installation drawing water from a reservoir.

Reference Plane

Fluid Machinery

573

determined from Equation 18.13 should be greater than the value given on a performance plot at the same flow rate.

Example 18.2 A centrifugal pump is needed to pump water from an open tank with a system configuration like that shown in Figure 18.10. The desired flow rate of water through a 10-cm-diameter pipe is 0.02 m3 /s, the manufacturer’s specifications show an NPSH value of 4.5 m at this flow rate. Determine the maximum height, z, above the water surface at which the pump should be placed to avoid cavitation. For this problem, the minor loss coefficient has a value of K L = 12 and all other losses may be neglected. Properties of water may be valuated at 27◦ C (300 K).

Solution Assumptions and Specifications (1) Steady flow exists. (2) The flow is incompressible. (3) There is uniform flow at the inlet and outlet sections. (4) Torques due to body and surface torques may be neglected. (5) Atmospheric pressure is taken as 101,360 Pa. From Equation 18.13, the available NPSH is NPSH =

 Patm Pv − z2 − hL − g g

which must be solved for z2 z2 ≤

Patm − Pv  − h L − NPSH g

At 27◦ C (300 K), water properties of interest include Pv = 3495 Pa

and

 = 997 kg/m3

 We are now able to evaluate Vˆ and h L ˙ V 4(0.02 m3 /s) Vˆ = = = 2.55 m/s A (0.10 m) 2 

hL = KL

Vˆ 2 12(2.55 m/s) 2 = = 3.98 m 2g 2(9.81 m/s2 )

so that the value of z2 can be determined as z2 ≤

101,360 Pa − 3495 Pa (997 kg/m3 )(9.81 m/s2 )

− 3.98 m − 4.5 m  1.526 m ⇐

The pump should be located no more than 1.526 m above the water level in the tank.

574

18.4

Introduction to Thermal and Fluid Engineering

Combining Pump and System Performance

In the preceeding discussion, we combined system characteristics with the realistic concern of avoiding cavitation, which will degrade pump performance and, eventually, lead to failure. We now examine the interaction between pump performance as described by operating curves of head, efficiency, and NPSH as a function of volumetric flow rate and the characteristics of the system in which the pump produces flow. Figure 18.11 illustrates a simple system configuration for a pump that produces flow between two reservoirs that are at different elevations. An energy balance between the surface of the lower reservoir, point 1, and the upper reservoir surface, point 2, gives ˙ W P2 − P1 (18.14) = g(z2 − z1 ) + + (u2 − u1 ) m ˙   Noting that P1 = P2 = Patm and writing u2 − u1 = g h L , we arrive at the head loss resulting from pipe friction and minor losses:    ˙ = mg −W ˙ z2 − z1 + hL −

or −

 ˙ W = z2 − z1 + hL mg ˙

(18.15)

We noted in Chapter 17 that h L is proportional to Vˆ 2 . Thus, h L = K Vˆ 2 and Equation 18.15 can be written as −

˙ W = z2 − z1 + K Vˆ 2 mg ˙

(18.16)

where K includes contributions from pipe friction and other minor losses due to items such as valves, elbows, and fittings. 2

pjwstk|402064|1435600812

1 z2

z1

FIGURE 18.11 Pump operating system.

Fluid Machinery

575

150

System Operating Line

100

11 η 10 9

50 NPSH

NPSH, m

Pump Efficiency, % Total Head, m

Head

8

0 0

0.50

1.0 1.50 Flow Rate, Vˆ m3/s

2.0

FIGURE 18.12 System operation and pump performance.

When the system operating line is plotted on the characteristic performance plot of the pump, the combined plot appears as shown in Figure 18.12. We note that the operating lines for the system and for the pump will intersect at a flow rate where the pumping head is equal to the required head needed for that flow rate to be achieved within the system. This operating point is associated with a corresponding pump efficiency as indicated in the figure. Naturally, the designer wants a given system operation to occur near the flow rate at which the pump efficiency is at a maximum. If such a match is not achieved, changes must be made either to the system (which is not likely to be feasible) or in the pump operating conditions. The scaling rules for pump performance will be considered in the next section. First, we work the following example where pump performance and system characteristics define a specific operating condition.

Example 18.3 The performance of a centrifugal pump is presented in Figure 8.12 for

operation at 1200 rpm. Water at 27◦ C (300 K) is to be pumped in the system shown in Figure 18.11. The two reservoirs differ in elevation by 40 m and the total length of 40-cm-diameter pipe is 480 m with a friction factor of f f = 0.0075. Determine the rate of flow for this system in m3 /s and evaluate the corresponding pump efficiency.

Solution Assumptions and Specifications (1) Steady flow exists. (2) The flow is incompressible. (3) There is uniform flow at the inlet and outlet sections. (4) Torques due to body and surface torques may be neglected. The energy equation for this system is −

˙S W L Vˆ 2 = z2 − z1 + 2 f f mg ˙ d g

576

Introduction to Thermal and Fluid Engineering

and when we insert the given numerical values, we obtain   

˙S W 480 m Vˆ 2 − = 40 m + 2(0.0075) mg ˙ 0.4 m 9.81 m/s2 or −

˙S W = 40 m + 1.835Vˆ 2 m mg ˙

˙ and we note that the For use with Figure 18.12, this expression must be in terms of V ˆ ˙ relationship between V and V is  2  d (0.4 m) 2 ˆ ˆ ˙ V=V =V = 0.1257Vˆ (m3 /s) 4 4 Substitution into the system energy balance expression yields  ˙ 2 ˙S W V − = 40 m + 1.835 m mg ˙ 0.1257 or −

˙S W ˙2m = 40 m + 116V mg ˙

The operating line corresponding to this expression is shown in Figure 18.12. The intersection of the pump and system operating lines is at a flow rate of ˙ = 0.90 m3 s ⇐ V where the efficiency is approximately  = 0.74 ⇐

18.5

Scaling Laws for Pumps and Fans

Concepts of similarity were introduced and demonstrated in Chapter 15. As discussed at that point, the requirements of geometric, kinematic, and dynamic similarity provide the basis for scaling, which is an extremely important process in predicting the performance of fluid machinery from laboratory- or bench-scale test results. The reader is referred to Chapter 15 for a review of the different types of similarity and how these concepts lead to predicting prototype performance. A dimensional analysis was performed for a centrifugal pump in Section 15.4. The relevant operating variables that were listed in Table 15.2 are repeated here for reference as Table 18.1. Then, by employing the formalism of the Buckingham method, the following dimensionless parameters were obtained ˙ gh V 1 = 2 2 3 = nd nd 3 P  2 = 3 5 4 = 2 n d d n It is easily demonstrated that these parameters are all dimensionless.

Fluid Machinery

577 TABLE 18.1

Pump Performance Variables: Power ˙ Is Designated as P Rather Than W Variable Total head Power Flow rate Impeller diameter Shaft speed Fluid density Fluid viscosity

Symbol

Dimensions

gh P ˙ V d n  

L 2 /T 2 ML 2 /T 3 L 3 /T L 1/T M/L 3 M/L T

The resulting parameters used in scaling pumps and fans are C H = the head coefficient,

gh n2 d 2

(18.17)

˙ V nd 3

(18.18)

P n3 d 5

(18.19)

C Q = the capacity coefficient, and C P = the power coefficient,

The relationships between these parameters provide the scaling laws we seek. For geometrically and kinematically similar pump families, dynamic similarity requires that C H = f 1 (C Q )

(18.20)

C P = f 2 (C Q )

(18.21)

and

We already know the other performance parameter as the efficiency, , which, by definition, is dimensionless. Using the coefficients just defined, we can write =

CHCQ = f 3 (C Q ) CP

(18.22)

and  is also seen to be a function of C Q . The relationships expressed in Equations 18.20 through 18.22 indicate that pump performance curves as introduced earlier can also be plotted using dimensionless quantities. Last, our three coefficients, C H , C Q , and C P , will provide the basis for scaling. Similarity requirements dictate that for two pumps that possess geometric, kinematic, and dynamic

578

Introduction to Thermal and Fluid Engineering

similarity, designated as pumps 1 and 2, the following relationships apply: C H1 = C H2 or gh 1 gh 2 = 2 2 2 2 n1 d1 n2 d2 so that h2 = h1



n2 n1

2 

d2 d1

2 (18.23)

Performing the same operations on C Q and C P , we obtain   ˙2 V n2 d2 3 = ˙1 n1 d1 V

(18.24)

and P2 2 = P1 1



n2 n1

3 

d2 d1

5 (18.25)

These three expressions comprise the pump laws or fan laws, which are universally used in scaling rotating machines and their performance. The following example demonstrates the utility of these scaling relationships.

Example 18.4 The pump described in Example 18.3 is operated at 1200 rpm against a head of 132 m H2 O. For a geometrically similar pump with a 20% larger impeller diameter, operating at the same rotational speed, what flow rate can be expected? In addition to the increase in the size of the larger pump, its speed is increased to 1400 rpm, determine the new values of flow rate, total head, and power required.

Solution Assumptions and Specifications (1) Steady flow exists. (2) The flow is incompressible. (3) There is uniform flow at the inlet and outlet sections. (4) Torques due to body and surface torques may be neglected. Designating pump 1 as the one specified in Example 18.3 and pump 2 with d2 = 1.2d1 , we have for n1 = n2 , ˙2 = V ˙ 1 n2 V n1



d2 d1

3

= (0.90 m3 /s)(1)(1.2) 3 = 1.555 m3 /s ⇐

Fluid Machinery

579

Then, for n2 = 1400 rpm and d2 = 1.2d1 , another application of Equation 18.24 provides  3 ˙2 = V ˙ 1 n2 d2 V n1 d1   1400 rpm = (0.90 m3 /s) (1.2) 3 1200 rpm = (0.90 m3 /s)(1.167)(1.2) 3 = 1.814 m3 /s ⇐ The total head is determined from Equation 18.23. With h 1 = 132 mH2 O,  2  2 n2 d2 h2 = h1 n1 d1   1400 rpm 2 = (132 mH2 O) (1.2) 2 1200 rpm = (132 mH2 O)(1.167) 2 (1.2) 2 = 258.7 mH2 O ⇐ The power ratio is given by Equation 18.25. With 2 = 1     P2 2 n2 3 d2 5 = P1 1 n1 d1   1400 rpm 3 = (1) (1.2) 5 1200 rpm = 3.95 The solution for P2 requires that we calculate P1 . The operating conditions for the original pump included ˙ 1 = 0.90 m3 /s V h 1 = 132 mH2 O  = 0.74 Neglecting minor losses for the time being, the power to the fluid is ˙ = PH2 O = mgh 1 = Vgh ˙ 1 −W and with  = 997 kg/m3 PH2 O = (997 kg/m3 )(0.90 m3 /s)(9.81 m/s2 )(132 m) = 1162 kW Then P1 =

PH2 O 1162 kW = = 1570 kW  0.74

and our final result for P2 is P2 = 3.95P1 = 3.95(1570 kW) = 6201 kW ⇐

580

18.6

Introduction to Thermal and Fluid Engineering

Axial and Mixed Flow Pumps

Our examination of pumps has, thus far, considered only centrifugal pumps. It was mentioned earlier that axial flow pumps are encountered in numerous applications. The distinction between axial and centrifugal flow is in the direction of fluid flow through the pump. In axial flow, the fluid moves parallel to the axis of rotation and in the centrifugal case, the flow is turned 90◦ normal to the rotational axis. It seems reasonable that there is an intermediate case in which the flow has both axial and normal components. This intermediate configuration is called mixed flow. The choice of a centrifugal, mixed or axial flow pump is determined by choosing an appropriate combination of head and volumetric flow rate that will achieve the peak efficiency for a given pump configuration. A useful parameter, in this regard, is obtained by eliminating the diameter between Q Q and C H . This operation yields the specific speed, NS , defined as

1/2

NS =

CQ

(18.26)

3/4

CH

Values of NS , at peak efficiency, for pump configurations ranging from centrifugal to axial are shown in the Figure 18.13 (lower axis). Also shown in the figure (upper axis) are values of NS expressed in what we call U. S. Customary Units. This parameter is dimensional and has numerical values determined using English-system dimensions as

˙ gpm]1/2 (, rpm)[V, [h, ft]3/4

Nsd =

(18.27)

Impeller Shrouds

20000 Axis of Rotation

Axial Flow

7.0 8.0

0.8 0.9 1.0

0.7

0.6

15000

Vanes

Mixed Flow 0.5

0.4

0.3

0.2

Radial Flow

Vanes

6.0

Vanes

Impeller Hub

5.0

Vanes

Hub Vanes

4.0

Hub

3.0

Hub

Impeller Shrouds

2.0

Hub

8000 9000 10000

Impeller Shrouds

7000

6000

5000

4000

3000

2000

1500

800 900 1000

700

600

500

Specific Speed, Nsd

Specific Speed, Ns FIGURE 18.13 Specific speed variation for various impeller configurations. (Adapted from Hydraulic Institute Standards, 14th ed., Hydraulic Institute, Cleveland, OH, 1983.)

Fluid Machinery

581

The basic message of this discussion, which is quantified in Figure 8.13, is that relatively large flow rates require the use of mixed flow and axial flow configurations.

18.7

Turbines

As stated at the beginning of this chapter, turbo machines either require power to produce flow and/or higher pressure or they produce power from a high-energy fluid. Machines that generate power are called turbines. Turbine operation involves a fluid that emanates from a nozzle, which interacts with blades attached to the periphery of a rotating unit, the rotor. The change in direction of the fluid flow, caused by the blades, results in a momentum exchange that produces power at the rotor shaft. The analysis of this momentum exchange process was examined in Chapter 14. Turbines are classified into two categories. In impulse turbines, there is no pressure drop across the rotor. In a reaction turbine, there is both a pressure drop and a change in the velocity of fluid flow through the rotor. Impulse turbines characteristically involve high head and low-flow rates and reaction turbines are low head, high-flow rate devices. A detailed discussion of turbines is beyond the scope of this book. The interested reader may consult any of a large number of treatises on this subject, some of which are cited in Fox and McDonald (1999), Munson et al. (1998), Potter and Wiggert (1997), Roberson and Grove (1997), and White (1999).

18.8

Summary

Turbo machines have been examined in this chapter. Included in our discussion and analysis have been those machines that require power to produce higher pressure, higher flow, or both. These machines are designated as pumps or fans. A machine that does the reverse, that is, produces power from a fluid, is designated a turbine. Pumps and fans are described in terms of the flow direction through the rotor. In centrifugal pumps (or fans), flow is turned 90◦ and in axial pumps (or fans), flow is along the rotational axis. A machine in which flow has both axial and radial components is designated as a mixed flow device. Each pump design is unique and is described by performance curves that are different in shape and operating values from other configurations. Operating variables of interest are pump head, flow rate, speed, characteristic size, power, and efficiency. A standard performance plot will show head, power, and efficiency as functions of the flow rate for a pump of characteristic impeller diameter at a particular speed of rotation. Scaling laws were obtained using dimensional analysis. For a family of geometrically similar pumps which also exhibit kinematic similarity, the scaling laws relate operating variables in the following ways: h2 = h1



n2 n1

2 

d2 d1

2

  ˙2 V n2 d2 3 = ˙1 n1 d1 V

(18.23)

(18.24)

582

Introduction to Thermal and Fluid Engineering

and P2 2 = P1 1

18.9



n2 n1

3 

d2 d1

5 (18.25)

Problems

Centrifugal Pumps 18.1: A centrifugal pump delivers 0.2 m3 /s of water when operating at 850 rpm. Pertinent impeller dimensions are as follows: outer diameter = 0.5 m, blade length = 50 cm, and blade exit angle = 25◦ . Determine (a) the torque and power required to drive the pump and (b) the maximum pressure increase across the pump. 18.2: A centrifugal pump is used with gasoline ( = 680 kg/m3 ). Pertinent dimensions are as follows: d1 = 15 cm, d2 = 28 cm, L = 9 cm, 1 = 25◦ , and 2 = 40◦ . The gasoline enters the pump parallel to the pump shaft when the pump operates at 1160 rpm. Determine (a) the flow rate in m3 /s, (b) the power delivered to the gasoline, and (c) the head in meters. 18.3: A centrifugal pump has the following dimensions: d2 = 45 cm, L = 5 cm, and 2 = 35◦ . It rotates at 1100 rpm. The head generated is 50 m of water. Assuming radial entry flow, determine the theoretical values for (a) the flow rate and (b) the power. 18.4: A centrifugal pump has the configuration and dimensions shown in Figure P18.4. For water flowing at a rate of 0.0071 m3 /s and an impeller speed of 980 rpm, determine the power required to drive the pump. The inlet flow is directed radially outward and the exiting velocity may be assumed to be tangent to the vane at its trailing edge. V = 0.0071 m3/s 55°

V1 980 rpm

28 cm 7.6 cm

1.90 cm FIGURE P18.4

18.5: A centrifugal pump is being used to pump water at a flow rate of 0.016 m3 /s and the required power is measured to be 4.4 kW. If the pump efficiency is 65%, determine the head generated by the pump. 18.6: A centrifugal pump having the dimensions shown in Figure P18.6, develops a flow rate of 0.028 m3 /s when pumping gasoline ( = 680 kg/m3 ). The inlet flow may be assumed to be radial. Estimate (a) the theoretical horsepower, (b) the head increase, and (c) the proper blade angle at the impeller inlet.

Fluid Machinery

583 5.1 cm 30° 10.2 cm 1750 rpm

6.5 cm

FIGURE P18.6

18.7: A centrifugal water pump operates at 1500 rpm. The dimensions are the following: r1 = 10 cm

1 = 30◦

r2 = 17.5 cm

2 = 20◦

L = 4.5 cm Determine (a) the design point discharge rate, (b) the water horsepower, and (c) the discharge head. 18.8: Figure P18.8 represents performance, in a nondimensional form, for a family of centrifugal pumps. For a pump from this family with a characteristic diameter of 55 cm operating at 1500 rpm pumping water at 15◦ C, operating at maximum efficiency, estimate (a) the head, (b) the discharge rate, (c) the pressure rise, and (d) the brake horsepower. 1.0 0.9 0.8 η

η 7

0.7 0.6

6

CH

5 CH

4 0.8 0.6 Cp

3

0.4 0.2

0

0.1

0.2 CQ

FIGURE P18.8

0.3

Cp

584

Introduction to Thermal and Fluid Engineering

18.9: We wish to build a pump having the characteristics of Figure P18.8 that will deliver water at a rate of 0.2 m3 /s when operating at best efficiency and a rotational speed of 1200 rpm. Estimate (a) the impeller diameter and (b) the maximum pressure rise. 18.10: Rework Problem 18.8 for a pump diameter of 35 cm operating at 2200 rpm. 18.11: Rework Problem 18.8 for a pump diameter of 30 cm operating at 2400 rpm. 18.12: Rework Problem 18.9 for desired flow rate of 0.25 m3 /s at 1800 rpm. 18.13: Rework Problem 18.9 for desired flow rate of 0.20 m3 /s at 1800 rpm. 18.14: A centrifugal pump with a 15-cm-diameter impeller has performance characteristics as indicated in Figure P18.14. The pump is to be used to pump gasoline ( = 680 kg/m3 and = 0.78 × 10−6 m/s2 ) through a 7.5-cm-diameter commercial steel pipe that has a length of 1200 m between two reservoirs at the same elevation. Determine the expected flow rate.

n = 3500 rpm 20 cm dia.

15 cm dia.

Head, m

100

15

10

50

5

0

NPSH, m

150

NPSH

0

0.2

0.4

0.6

0.8

1.0

1.2

0 1.4

Capacity, m3/m FIGURE P18.14

18.15: Determine the flow rate for the system described by Problem 18.14 if the pipe diameter is increased to 10 cm. 18.16: Rework Problem 18.14 for an impeller that has a diameter of 16.5 cm. 18.17: Rework Problem 18.14 if the impeller diameter is increased to 18 cm. Scaling Laws 18.18: Performance curves for an operating centrifugal pump are shown in Figure P18.18a and in a dimensionless form in Figure P18.18b. This pump is used to pump water at maximum efficiency at a head of 80 m. Determine, at these new conditions, (a) the pump speed required and (b) the rate of discharge.

Fluid Machinery

585 100

250

80

Efficiency

Power Input, kW

150

60 Head (ΔH), m 40

100

n = 360 rpm d = 39 cm

50

0

0

0.05

0.10

Efficiency, %

Head, m; Power, kW

200

20

0.15

0.20

0.25

Discharge, m3/s (a)

6 CH 1.00

5

CH

0.75 CP

3

0.50

CP and Efficiency

Efficiency-η 4

2 0.25

1

0

0

0.04

0.08 CQ

0.12

0 0.16

(b)

FIGURE P18.18

18.19: The pump having the characteristics shown in Figure P18.18 was used as a model for a prototype that is to be 6 times larger. If this prototype operates at 400 rpm, what (a) power, (b) head, and (c) discharge flow rate should be expected at maximum efficiency? 18.20: For the pump having the characteristics shown in Figure P18.18 operating at the conditions specified with the speed increased to 1000 rpm, what will be the (a) new discharge flow rate and (b) the power required at this new speed? 18.21: The pump having the characteristics shown in Figure P18.18 is to be operated at 750 rpm. What discharge rate is to be expected if the head developed is 424 m? 18.22: If the pump having the characteristics shown in Figure P18.18 is tripled in size but halved in rotational speed, what will be the discharge rate and head when operating at maximum efficiency? 18.23: Plot the head-discharge curve for a centrifugal pump geometrically similar to the one having the characteristics shown in Figure P18.18 but with a diameter of 1.6 m operating at 650 rpm.

586

Introduction to Thermal and Fluid Engineering

18.24: A 12.5-cm-diameter centrifugal pump operating at 1800 rpm displays the following characteristics when pumping water: ˙ m3 /s V, h, m , percent

0.03 10.6 18

0.09 11.9 66

0.15 12.0 84

0.21 10.7 73

0.27 7.3 25

Determine the power required when this pump runs at 2400 rpm and delivers a flow of 0.170 m3 /s. 18.25: A pump that is geometrically similar to the one specified in Problem 18.24 is to deliver 0.027 m3 /s when operating at 3000 rpm, determine the power required. 18.26: Performance curves for an axial flow pump are shown in Figure P18.26a and, in a dimensionless form, in Figure P18.26b. When this pump operates at 690 rpm in a system requiring 3 m of head, what flow rate will be produced and what is the required power? 100

Efficiency

80

60

Head

5

40

20 n = 690 rpm n = 11.5 rps d = 35.6 cm 0

0.05

0.15

0.10

0.20

0.25

0.30

Discharge, m3/s (a)

100

5.00 η 4.00

80

3.00

60

2.00

CP

1.00

0

40

CH

0

0.2

20

0.4 CQ

0.6

(b)

FIGURE P18.26

0 0.8

Efficiency, %

0

Efficiency, %

Power

Cp and CH

Head, m; Power, kW

10

Fluid Machinery

587

18.27: A prototype axial flow pump is geometrically similar to the one whose operating characteristics are shown in Figure P18.26a and P18.26b. The pump has a characteristic diameter of 36 cm and is to operate at 900 rpm to produce a volumetric flow rate of 1.5 m3 /s. Determine (a) the head generated and (b) the level of power required. 18.28: A pump, which is geometrically similar to the one whose operating characteristics are shown in Figure P18.26a, is operated at 720 rpm and has a characteristic diameter of 80 cm. Determine (a) the discharge rate and (b) power required when the head is 150 m. 18.29: A pump that is geometrically similar to the one whose operating characteristics are shown in Figure P18.26 is to have a characteristic diameter of 65 cm. Determine (a) the discharge rate, (b) the head produced, and (c) the required power when the pump is operating at maximum efficiency at 900 rpm. 18.30: Rework Problem 18.7 for a centrifugal pump that is geometrically similar to the pump specified but 50% larger in its physical dimensions. The larger pump will operate at 1500 rpm. 18.31: Rework Problem 18.7 with all specifications remaining the same except for a new rotational speed of 1220 rpm. 18.32: Rework Problem 18.7 with all specifications remaining the same except for a new rotational speed of 1850 rpm. 18.33: A centrifugal pump with an impeller diameter of 1.2 m is being constructed to provide a flow rate of 4.5 m3 /s of water with a head rise of 180 m when it operates at 1400 rpm. A geometrically similar model at 1/8 scale is to be tested in the laboratory. Assuming both model and prototype to operate at the same rotational speed, determine (a) the model discharge flow rate and (b) the model head rise. 18.34: Determine and plot values for the dimensionless coefficients, C H , C P , C Q , and  for the pump data given in Problem 18.24. 18.35: A centrifugal pump with a 30-cm-diameter impeller requires input power of 45 kW when producing 0.20 m3 /s of flow against a head of 61 m. Determine (a) the expected flow rate, (b) head, and (c) the power if the pump operates at the same speed as an impeller having a diameter of 25 cm. 18.36: A pump, which is geometrically similar to the one whose operating characteristics are shown in Figure P18.26, has a characteristic diameter of 70 cm. It is desired to operate this pump at 900 rpm and at maximum efficiency. Determine (a) the discharge rate, (b) the head, and (c) the power required. Pump/System Compatability 18.37: The pump having the characteristics shown in Figure P18.18 is used to pump water from one reservoir to another that is 90 m higher in elevation. The water will flow through a commercial steel pipe, 32 cm in diameter and 520 m in length. Determine the discharge rate. 18.38: A pump whose operating characteristics are shown in Figure P18.26 is to be used in the system in Figure P18.38. Determine (a) the discharge rate and (b) power required.

588

Introduction to Thermal and Fluid Engineering Diameter = 36 cm 90 m Steel Pipe

Elevation = 24 m

0.5 m

Elevation = 18 m 3m

FIGURE P18.38

18.39: For the same pump and system operation described in Problem 18.38, determine (a) the discharge rate and (b) power required when the pump operates at 900 rpm. 18.40: Water at 20◦ C is to be pumped through the system shown in Figure P18.40. The operating data for a motor-driven pump are as follows: Capacity, m3 /s

Developed Head, m

Efficiency, %

0 10 20 30 40 50

36.6 35.9 34.1 31.2 27.5 23.3

0 19.1 32.9 41.6 42.2 39.7

The inlet pipe to the pump is commercial steel, 6.5 cm in diameter and 7.6 m in length. The discharge line consists of 64 m of 16.5-cm-diameter steel pipe and all valves are fully open globe valves. Determine the flow rate through the system and the electrical power necessary to achieve this flow.

18 m

3m FIGURE P18.40

Net Positive Suction Head 18.41: A 20-cm pump delivers 20◦ C water (Pv = 2.34 kPa) at 0.0631 m3 /s and 2200 rpm. The pump begins to cavitate when the inlet pressure is 82.7 kPa and the inlet velocity is 6.1 m/s. Determine the corresponding NPSH. 18.42: For the pumping system described in Problem 18.41, how will the maximum elevation above the surface of the reservoir change if the water temperature is 80◦ C (Pv = 47.35 kPa)?

Fluid Machinery

589

18.43: The performance curves for a centrifugal pump with an impeller diameter of 20 cm are shown in Figure P18.43. This pump is to be used to pump water ( = 1000 kg/m3 ) with the pump inlet located 3.5 m above the surface of the supply reservoir. At a flow rate of 0.760 m3 /s, the head loss between the reservoir surface and the pump inlet is 1.83 m of water. Would you expect cavitation to occur? n = 3300 rpm

20-cm diameter

15-cm diameter

100

15

10

50

0

NPSH, m

Head, m

150

5 NPSH 0

0.2

0.4

0.8 0.6 Capacity, m3/s

1.0

1.2

0 1.4

FIGURE P18.43

18.44: The pump whose performance is given in Figure P18.43 is to be placed above a large open tank and is to pump water at a rate of 0.2 m3 /s. The system head loss can be assumed to have a coefficient of K = 0.3 and the pump inlet has a diameter of 12 cm. Determine the maximum height at which the pump can be located without experiencing cavitation. 18.45: A prototype pump, geometrically similar to the one specified in Problem 18.41, is 5 times larger and runs at 1200 rpm. What is the required NPSH for the prototype? 18.46: A centrifugal pump is installed z meters above a reservoir. The manufacturer’s specifications indicate the value of NPSH to be 4.8 m when the system is pumping water at 20◦ C (Pv = 2.34 kPa) at a rate of 0.014 m3 /s. All losses in the installation combine to yield a loss coefficient of K L = 20 and the inlet pipe has a diameter of 10 cm. Determine the maximum elevation, z, above the reservoir surface at which the pump can be positioned without experiencing cavitation. Specific Speed 18.47: Pumps used in an aqueduct operate at 400 rpm and deliver a flow of 200 m3 /s against a head of 450 m. What types of pumps are they? 18.48: A pump is required to deliver 60,000 gpm against a head of 290 m when operating at 2000 rpm. What type of pump should be specified? 18.49: An axial flow pump has a specified specific speed of 5.0. The pump must deliver 2500 gpm against a head of 16 m. Determine the required operating rpm of the pump. 18.50: A pump operating at 400 rpm has the capability of producing 3 m3 /s of water flow against a head of 18 m. What type of pump is this? 18.51: A pump operating at 1800 rpm produces a head rise of 27 m and a discharge flow rate of 0.045 m3 /s when the power required is 18.5 kW. Determine (a) the specific

590

Introduction to Thermal and Fluid Engineering speed for this pump and the new values of (b) head rise and (c) shaft horsepower if the pump speed is decreased to 1400 rpm.

18.52: A centrifugal pump achieves a discharge flow rate of 0.032 m3 /s when operating at 1800 rpm against a head of 60 m. Determine the new (a) flow rate and (b) developed head if the rotational speed is increased to 3600 rpm. 18.53: It has been determined that, in a specific application, a pump must deliver 0.32 m3 /s of flow against a head of 90 m when operating at 1200 rpm. What type of pump would you recommend? 18.54: A pump operating at 2160 rpm delivers 3.5 m3 /s of water against a head of 20 m. Is this pump an axial flow, mixed flow, or radial flow machine? 18.55: An axial flow pump with a specific speed of 6.0 is required to deliver 0.20 m3 /s of water against a head of 4.5 m. At what speed should this pump be operated?

19 Introduction to Heat Transfer

Chapter Objectives To introduce and describe the three modes of heat transfer: conduction, convection, and radiation. • To define thermal conductivity, a heat transfer property. •

•

To develop the concept of thermal resistance.

•

To consider the overall heat transfer coefficient.

19.1

Introduction

pjwstk|402064|1435600821

The remaining eight chapters of our study deal with heat transfer. The quantity, heat, was defined in earlier chapters and it was quantitatively related to work and system behavior through the first law of thermodynamics in Chapter 5. The first law remains a fundamental precept as we now devote our attention to the rate of heat exchange. As discussed earlier, heat will be exchanged between systems when they are at different temperatures. The second law of thermodynamics stipulates that this exchange will be from the higher-temperature body or system, toward the bodies or systems at lower temperatures. Practical considerations regarding heat transfer processes involve the rate at which heat transfer occurs. All decisions that involve equipment and material specifications require that heat transfer rates and process temperatures be determined. Our goal, in this chapter, is to examine the mechanisms of heat transfer and to introduce the equations that are fundamental to the evaluation of rates at which energy transfer, due to temperature difference, occurs.

19.2

Conduction

Energy transfer by conduction is accomplished via two mechanisms. The first mechanism is by molecular interaction, in which the greater motion of a molecule at a higher-energy (temperature) level imparts energy to adjacent molecules at lower-energy levels. This type of energy transfer is present, to some degree, in all systems in which a temperature gradient exists and in which molecules of a solid, liquid, or gas are present. The second mechanism of conduction heat transfer is by “free” electrons. The free-electron mechanism is significantly greater in pure metallic solids; the concentration of free electrons 591

592

Introduction to Thermal and Fluid Engineering

varies considerably for alloys and becomes very low for nonmetallic solids. The ability of solids to conduct heat varies directly with the concentration of free electrons. Thus, it is not surprising that pure metals are the best heat conductors, as our experience has indicated. Because heat conduction is primarily a molecular phenomenon, we might expect that the basic equation used to describe this process is similar to the expression used in the molecular transfer of momentum, Equation 11.19. Such an equation was first stated in 1822 by Fourier in the form q˙ x =

˙x Q dT = −k A dx

(19.1)

˙ x , per unit of area normal where q˙ x is the heat flux, defined as the rate of heat transfer, Q to the direction of heat flow, dT/d x is the temperature gradient, and k is the thermal conductivity in W/m-K. A more general relation for the heat flow is q˙ =

˙ Q = −k∇T A

(19.2)

which expresses the heat flux (and rate of heat transfer) as proportional to the temperature gradient. The proportionality constant is the thermal conductivity that plays a role similar to that of viscosity in momentum transfer. The negative signs in both Equations 19.1 and 19.2 indicate that the heat flow is in the direction of a negative temperature gradient. Equation 19.2 is the vector form of the Fourier rate equation and is often referred to as Fourier’s law of heat conduction. The thermal conductivity, k, which is defined by Equation 19.1 is assumed independent of direction in Equation 19.2. Thus, Equation 19.2 applies only to an isotropic medium. The thermal conductivity is a property of a conducting medium and, like the viscosity, is primarily a function of temperature, varying significantly with pressure only in the case of gases subjected to high pressures. Under steady-state conditions, Equation 19.1 can be solved to evaluate the heat transfer by conduction between two locations. Its solution is achieved by the method of separation of the variables and invoking the boundary conditions where T(x = x1 ) = T1 , T(x = x2 ) = T2 , to yield ˙ = k A(T1 − T2 ) Q L

(19.3)

where L = x2 − x1 . For one-dimensional, steady, heat flow in cylindrical coordinates where the heat flow is in the radial direction, Equation 19.1 becomes ˙ = −k AdT = −k(2r L) dT Q dr dr

(19.4)

where A = 2r L . For one-dimensional, steady, heat flow in spherical coordinates where the heat flow is in the radial direction, Equation 19.1 becomes ˙ = −k AdT == −k(4r 2 ) dT Q dr dr

(19.5)

where A = 4r 2 . Equations 19.4 and 19.5 may be solved using separation of the variables subject to the boundary conditions T(r = r1 ) = T1

and

T(r = r2 ) = T2

Introduction to Heat Transfer

593

to yield for the cylinder ˙ = 2k L(T1 − T2 ) Q ln r2 /r1

(19.6)

and for the sphere ˙ = 4k(T1 − T2 ) Q 1 1 − r1 r2 With 1 1 r2 − r1 − = r1 r2 r1 r2 we have ˙ = 4kr1r2 (T1 − T2 ) Q r2 − r1

19.3

(19.7)

Thermal Conductivity

Equation 19.1 is the defining equation for the thermal conductivity and experimental measurements made on the basis of this definition may be employed to evaluate the thermal conductivities of different materials. In the solid phase, thermal conductivity is attributed to both molecular interaction and free-electron drift that is present, primarily, in pure metals. The solid phase is amenable to quite precise measurements of thermal conductivity because there is no effect of convection currents. The thermal properties of most solids of engineering interest have been evaluated. Table A.17 (for metals) and Table A.18 (for nonmetallic solids) listing thermal conductivities and other properties are included in Appendix A. The thermal conductivity of a liquid is not amenable to any simplified kinetic-theory development because the molecular behavior of the liquid phase is not clearly understood and no universally accurate mathematical model presently exists. Some empirical correlations have met with reasonable success, but they are so specialized they will not be included in this book. A general observation about liquid thermal conductivities is that they vary only slightly with temperature and are relatively independent of pressure. One problem in experimentally determining values of the thermal conductivity in a liquid is making sure that the liquid is free of convection currents. Table A.14 gives the thermal conductivity of water as well as other properties and Table A.16 provides similar properties for several other liquids. The kinetic theory of gases can be used to predict the thermal conductivity of gases and experiments confirm that the thermal conductivity is proportional to the absolute temperature, Table A.13 gives the thermal conductivity of air as well as other properties and Table A.15 provides similar properties for several other gases. Figure 19.1 illustrates the thermal conductivity variation with temperature of several important materials in solid, liquid, and gas phases.

Example 19.1 As indicated in Figure 19.2, a pure aluminum cylinder having an inside diameter of 1.880 cm and a wall thickness of 0.391 cm is exposed to inside and outside surface temperatures of 94◦ C and 71◦ C. Determine, per meter of pipe length (a) the heat flow rate and (b) the heat fluxes based on both the inside and outside surface area.

Introduction to Thermal and Fluid Engineering

k, W/m-°C

594

Copper

400 300

Pure Aluminum

200 5% of Steel

100 0

20

200 Temperature, °C (a)

300

Glycerin

0.3

k, W/m-°C

100

Ethylene Glycol 0.2 Unused Engine Oil

0.1

0.0 0

10

20

30

40 50 60 Temperature, °C (b)

70

80

0.05

k, W/m-k

0.04 0.03 O2

0.02 CO2

0.01 0.00 100

200

300 400 Temperature, K

500

600

(c) FIGURE 19.1 Representative thermal conductivities of (a) metals, (b) liquids, and (c) gases as functions of temperature.

Solution Assumptions and Specifications 1. The heat flow is steady. 2. The heat flow is one dimensional.

Introduction to Heat Transfer

595

71°C 94°C

0.391 cm

1.880 cm FIGURE 19.2 Configuration for Example 9.1.

Here, r1 and r2 designate the inner and outer radii, respectively, and because the heat flow is in the radial direction, Equation 19.4 applies ˙ = −k AdT = −k(2r L) dT Q dr dr With the variables separated this becomes ˙ dr = −2kLdT Q r and the integration gives r2 r2 T2    dr   ˙ ˙ ln r = Q ˙ ln r2 /r1 = −k(2L)T  = 2k L(T1 − T2 ) Q =Q    r r1 r1 T1 or finally ˙ = 2k L(T1 − T2 ) = 2k L(T1 − T2 ) Q ln r2 /r1 ln d2 /d1

(19.6)

Notice that the radius ratio r2 /r1 can be replaced by the diameter ratio, d2 /d1 . Here, the radius ratio will be used and with di = d1 = 1.880 cm

and

do = d2 = 1.880 cm + 2(0.391 cm) = 2.662 cm

These make r1 = 0.94 cm

and r2 = 1.331 cm

The arithmetic average temperature in the pipe wall is Tav =

94◦ C + 71◦ C 165◦ C = = 82.5◦ C 2 2

Table A.17 reveals (with interpolation) that the thermal conductivity of pure aluminum at Tav = 82.5◦ C is k = 205.6 W/m-K. (a) The heat flow will be ◦ ◦ ˙ = 2(205.6 W/m-K)(1.00 m)(94 C − 71 C) = 26, 712 W = 85.43 kW ⇐ Q ln(0.01331 m/0.0094 m) 0.3478

596

Introduction to Thermal and Fluid Engineering

(b) The heat flux varies as a function of location. At the inner surface q˙ =

85.43 kW = 1446 kW/m2 ⇐ 2(0.0094 m)(1.00 m)

and at the outer surface q˙ =

85.43 kW = 1021.5 kW/m2 ⇐ 2(0.01331 m)(1.00 m)

Notice that for the same amount of heat flow, the heat fluxes based on the inside and outside areas of the pipe are vastly different, differing by about 42%.

19.4

Convection

Heat transfer by convection occurs between a surface and an adjacent fluid. When the fluid is induced to flow past the surface by an external agency such as a pump or fan, the condition is that of forced convection. When fluid motion is induced by density differences in the fluid caused by the heating or cooling of the fluid through the heat exchange process, the process is called natural convection or free convection. The rate equation for convective heat transfer is attributed to Isaac Newton (1701). The Newton rate equation, sometimes referred to as “Newton’s law of cooling,” is q˙ =

˙ Q = hT S

(19.8)

˙ is the rate of heat transfer in W, S is the surface area,1 and h is the coefficient of heat where Q transfer in W/m2 -K. Equation 19.8 is, fundamentally, the defining relationship of h. One of the principal challenges in quantifying convective heat transfer is that of determining the appropriate value for this quantity. The coefficient, h, is a function of the system geometry, fluid and flow properties, and the magnitude of the temperature difference, T. Because fluid behavior is involved in convective heat transfer, it will be necessary to use our knowledge of fluid mechanics considered in earlier chapters of this text. We will need to characterize the flow as laminar or turbulent, and will have occasion to use many of the concepts and methods of analysis developed previously. Heat transfer involving a change of phase, while not technically convective processes, are still treated quantitatively using Equation 19.8. When either boiling or condensation occur, relatively high rates of heat transfer are involved even though the magnitude of T may be relatively small. Table 19.1 lists values of h typical of five different convective mechanisms.

19.5

Radiation

Radiant heat transfer differs from conduction and convection in that no medium is required for its propagation. Indeed, energy transfer by radiation is at maximum when the two surfaces that are exchanging energy are separated by a perfect vacuum. 1 In

this book, a distinction is made between cross-sectional area, A, and surface area, S.

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TABLE 19.1

Approximate Values of the Convection Heat Transfer Coefficient h, W/m2 -K

Mechanism Free convection, air Forced convection, air Forced convection, water Boiling water Condensing water vapor

5–50 25–250 250–15,000 2500–25,000 5000–100.000

The rate of energy emission from a perfect radiator or blackbody is given by ˙ = ST 4 Q

(19.9)

where  is the Stefan-Boltzmann constant  = 5.67 × 10−8 W/m2 -K4 ˙ is the rate of radiant energy transfer in W, S is the surface area of the In Equation 19.9, Q emitting surface, and T is the absolute temperature of the emitting surface in K. The pro˙ portionality constant, , relating the radiant energy flux, q˙ = Q/S, to the fourth power of the absolute temperature is named after Stefan who, from experimental observations, proposed Equation 19.9 in 1879, and Boltzmann, who derived this equation theoretically in 1884. Equation 19.9 is most often referred to as the Stefan-Boltzmann law of thermal radiation. Certain modifications will be made to Equation 19.9 to account for the net energy transfer between two surfaces, the degree of departure of the emitting surface and the receiving surface from ideal emitter and receiver behavior, and geometric factors associated with radiant interchange between a surface and its surroundings. If the surface and geometric effects are lumped into a factor F, then, we may write for the heat transfer by radiation between surface 1 at T1 to surface 2 at T2   ˙ = SF T14 − T24 Q

19.6

(19.10)

Thermal Resistance

The steady-state heat flow in one dimension through a plane wall is represented by Equation 19.4 ˙ = k A(T1 − T2 ) Q L

(19.4)

where L is the wall thickness and T1 and T2 are the surface temperatures. This equation may be put into the form of a thermal resistance Rcond ≡

T1 − T2 T L = = ˙ ˙ kA Q Q

(19.11a)

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Introduction to Thermal and Fluid Engineering

which, not only defines a thermal resistance in conduction, but also suggests an electrothermal analogy with Ohm’s law R=

V I

in which V ⇐⇒ T ˙ I ⇐⇒ Q R ⇐⇒ Rthermal The thermal resistances of the cylinder and the sphere in conduction are respectively, Rcond =

ln r2 /r1 2k L

(19.11b)

Rcond =

r2 − r1 4kr1r2

(19.11c)

and

Using this same reasoning, Equation 19.8, written in the form, ˙ = h ST Q becomes, upon rearrangement, Rconv =

T 1 = ˙ hS Q

(19.12)

For the case of radiation, Equation 19.8   ˙ = F S T14 − T24 Q

(19.8)

the thermal resistance is Rrad =

1 h rad S

(19.13)

in which   h rad = F T12 + T22 (T1 + T2 )

19.7

(19.14)

Combined Mechanisms of Heat Transfer

The three heat transfer modes have been introduced in Sections 19.2, 19.4, and 19.5. Most often more than one of these mechanisms occur in combination. In this section, we examine some typical situations where more than one heat transfer mode is involved. We first consider the plane wall shown in Figure 19.3, with steady-state conduction occurring through the wall whose surfaces are at temperatures T1 and T4 , respectively. The wall consists of three different materials, having thermal conductivities, k2 through k4 and

Introduction to Heat Transfer

599

k2

T1 T2

k3

k4

h5

T3

h1

T5

T4

T6

FIGURE 19.3 Composite plane wall with convection at its faces.

each external face of the wall is exposed to a convective environment. At the left face, the environmental temperature is T1 and the heat transfer coefficient is h 1 and at the right face, the environmental temperature is T6 and the heat transfer coefficient is h 6 . Observe that there are five temperature differences of interest and that each of them is related to the common heat flow through them via an appropriate thermal resistance:

˙ 1=Q ˙ T1 − T2 = QR ˙ 2=Q ˙ T2 − T3 = QR ˙ 3=Q ˙ T3 − T4 = QR

  



L2 k2 A L3 k3 A

 

 L4 k4 A   1 ˙ 5=Q ˙ T5 − T6 = QR h5 S ˙ 4=Q ˙ T4 − T5 = QR



1 h1 S

Adding these yields

pjwstk|402064|1435600826

˙ T1 − T6 = Q



1 L2 L3 L4 1 + + + + h 1 S k2 A k3 A k4 A h 5 S

and because A = S for the plane wall, this is equivalent to

T1 − T6 =

 ˙ 1 Q L2 L3 L4 1 + + + + S h1 k2 k3 k4 h5

The electrothermal analog is shown in Figure 19.4.

 (19.15)

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Introduction to Thermal and Fluid Engineering

T2

L2 k2A

T3

L3 k3A

T4

L4 k4A

1 h1S

T1 – T6

T5

1 h5S

+ – T6

FIGURE 19.4 Electrothermal analog network for the plane wall of Figure 19.3.

Example 19.2 Saturated steam at 0.276 MPa flows inside of a steel pipe having a thermal

conductivity of k = 42.9 W/m-K, an inside diameter of 2.09 cm, and an outside diameter of 2.67 cm. The convective coefficients on the inner and outer pipe surfaces may be taken as 5680 W/m2 -K and 22.7 W/m2 -K, respectively. The surrounding air is at 21◦ C. Find the heat loss per meter of bare pipe.

Solution Assumptions and Specifications 1. The heat flow is steady. 2. The heat flow is one dimensional. Here, the convective resistances are designated as R1 at the inside and R3 at the outside of the pipe. The conduction resistance for the pipe wall is designated as R2 . These three thermal resistances are evaluated first. With ri and ro taken as the inside and outside radii, respectively R1 = = R2 = = R3 = =

1 1 = h i Si 2h i ri L 1 2

2(5680 W/m -K)(0.01045 m)(1.00 m)

= 0.00268 K/W

ln r2 /r1 2k L ln(1.335 cm/1.045 cm) = 0.00091 K/W 2(42.9 W/m-K)(1.00 m) 1 1 = h o So 2h o ro L 1 2

2(22.7 W/m -K)(0.01335 m)(1.00 m)

= 0.5252 K/W

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The total thermal resistance is RT = R1 + R2 + R3 = 0.00268 K/W + 0.00091 K/W + 0.5252 K/W = 0.5288 K/W Table A.4 (with interpolation) gives the saturation temperature of steam at 276 kPa as 131◦ C so that the heat loss per meter of pipe length is ◦ ◦ ˙ = T = 131 C − 21 C = 208 W ⇐ Q RT 0.5288 K/W

19.8

The Overall Heat Transfer Coefficient

An alternative means for treating a heat transfer situation involving combined modes is through the use of the overall heat transfer coefficient, U. This quantity is defined according to the relationship ˙ = U ST Q

(19.16)

where U accounts for all of the thermal resistances between the two extreme temperatures. Equation 19.16 is straightforward when a plane composite wall is involved but when the system is cylindrical or spherical, the value of U will vary depending on the choice of the reference area, S. For the case with cylindrical geometry, as in Example 19.2, we could use either of the forms ˙ = Uo So T Q

or

˙ = Ui Si T Q

(19.17)

In the two equivalent forms given by Equation 19.17, the coefficients , Uo and Ui , are associated with outside (outer) and inside (inner) surface areas, respectively. In Example 19.2, the total resistance for the three resistance configuration could have been written as RT =

1 ln r2 /r1 1 + + h i Si 2k L h o So

Suppose that the outer surface is taken as the reference. Then Uo =

˙ Q T/RT 1 = = So T So T So RT

and Uo = 

1  So So ln r2 /r1 1 + + h i Si 2k L ho

(19.18)

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Introduction to Thermal and Fluid Engineering

and if the inner surface is taken as the reference Ui =

˙ Q T/RT 1 = = Si T Si T Si RT

and Ui = 

1  1 Si ln r2 /r1 Si + + hi 2k L h o So

(19.19)

Example 19.3 Rework Example 19.2 using the overall heat transfer coefficient based on the outside surface.

Solution Assumptions and Specifications 1. The heat flow is steady. 2. The heat flow is one dimensional. Here, So = 2ro L = 2(0.01335 m)(1.00 m) = 0.0839 m2 and Si = 2ri L = 2(0.01045 m)(1.00 m) = 0.0657 m2 With Uo =

1 So RT

and Ui =

1 Si RT

we can take RT from Example 9.2 RT = 0.5288 K/W and obtain Uo =

1 1 = = 22.54 W/m2 -K 2 So RT (0.0839 m )(0.5288 K/W)

Ui =

1 1 = = 28.80 W/m2 -K 2 Si RT (0.0657 m )(0.5288 K/W)

and

Thus, ˙ = Uo So T = (22.54 W/m2 -K)((0.0839 m2 )(110◦ C) = 208 W ⇐ Q and ˙ = Ui Si T = (28.78 W/m2 -K)((0.0657 m2 )(110◦ C) = 208 W ⇐ Q

Introduction to Heat Transfer

19.9

603

Summary

The rate equations for heat transfer in the steady state and in one dimension are as follows: For conduction in rectangular coordinates ˙ x = −k AdT Q dx

(19.1)

and for conduction in cylindrical and spherical coordinates ˙ r = −k AdT Q dr

(19.4, 19.5)

For convection ˙ = h ST Q

(19.8)

˙ = ST 4 Q

(19.10)

and for radiation

Each mode may be represented by a thermal resistance: L kA

(19.11a)

Rcond =

ln r2 /r1 2k L

(19.11b)

Rcond =

r2 − r1 4kr1r2

(19.11c)

Rcond = for plane surfaces

for cylinders and

for spheres. In addition, thermal resistances exist for convection and radiation 1 hS

(19.12)

1 h rad S

(19.13)

Rconv = and Rrad = in which

  h rad = F T12 + T22 (T1 + T2 )

(19.14)

The overall heat transfer coefficient is designated as U and must be linked to a particular surface area. When this is done, ˙ = Uo So T = Ui Si T Q

(19.16)

604

19.10

Introduction to Thermal and Fluid Engineering

Problems

Conduction 19.1: Heat flows at a rate of 4 kW through an insulating material having a cross-sectional area of 12 m2 and a thickness of 3.6 cm. The hot surface is at 112◦ C and the thermal conductivity of the material is 0.25 W/m-K. Determine the cold surface temperature. 19.2: A temperature difference of 80◦ C exists across a 12-cm layer of insulating material having a thermal conductivity of 0.032 W/m-K. Determine the heat transferred per unit area. 19.3: An igloo in the shape of a perfect hemisphere, has an outside diameter of 4 m and the ice walls are 18-cm thick. If the inner face of the ice is at 0◦ C and the outer face is at −12◦ C, determine the heat transferred through the wall of the igloo. 19.4: A pyrex glass beaker with a thermal conductivity of 0.56 W/m-K has an outside diameter of 12.70 cm, a height of 15.24 cm, and walls that are 3.5-mm thick. The beaker contains a hot liquid such that the inside surface is maintained at 92◦ C. If the outside surface is held at 17◦ C, determine the heat transferred through the walls of the beaker. 19.5: A beverage carrier has inside dimensions of 50 cm × 25 cm and is 20-cm high. It is fabricated of a material having a thermal conductivity of 0.023 W/m-K. The walls of the carrier are 2.25-cm thick and the outside temperature of the carrier is at 28◦ C. Suppose that the latent heat of fusion of ice is 333.7 kJ/kg and that the entire inside volume of the carrier is filled with ice. Neglect the heat flow through the base of the carrier and determine the amount of ice that will melt in 1 hour. 19.6: The inside surface of an 8 in, schedule 80, 0.5% carbon steel pipe is maintained at 39◦ C and the outside surface is held at 21◦ C. Determine the heat loss per meter of length. 19.7: It is proposed to use Teflon to coat a 2-in-OD tube. If the surface of the tube is held at 82◦ C and the surface of the Teflon is maintained at 18◦ C, determine the thickness of the Teflon needed to limit the heat loss to 40 W/m of tube length. 19.8: A basement has a dry concrete floor that is 10-m long, 8-m wide, and 20.32-cm thick. The top and bottom temperatures of the floor are 17◦ C and 8◦ C, respectively. The basement is heated by a gas furnace and the gas costs $0.08/MJ. Consider the heat loss through the floor and determine the hourly cost of the gas. Convection 19.9: A wall that is 8-m high and 6-m wide is held at 37◦ C. If the heat transfer coefficient between the wall and the surroundings is 7.2 W/m2 -K and the surroundings are at 17◦ C, determine the heat loss by convection. 19.10: A small 4-cm-diameter sphere is maintained at 62◦ C in an enclosure where the air is at 44◦ C. If the heat transfer coefficient is 14 W/m2 -K, determine the heat loss by convection. 19.11: How much heat will be transferred by convection from an 8-m long 3-in schedule 80 pipe held at 77◦ C in a room held at 17◦ C where the heat transfer coefficient is 10.25 W/m2 -K?

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605

19.12: An electric heater is embedded in a 3-in tube. When the outside surface of the tube is held at 80◦ C and the air surrounding the tube is at 16◦ C, it is noted that 640 W is provided to the heater. If the tube is 2-m long, determine the heat transfer coefficient. 19.13: A power transistor may be treated as a capped cylinder having an outside diameter of 12.5 mm and a height of 15 mm. The transistor is cooled by air at 37◦ C via a heat transfer coefficient of 75 W/m2 -K and its case is to be held at 72◦ C to satisfy reliability constraints. Determine the power rating of the transistor for these conditions. 19.14: Consider a square microcircuit chip that is 4 mm on a side and mounted in an insulated substrate. If the surface of the chip is to be held at 77◦ C because of reliability constraints and air at 17◦ C under a heat transfer coefficient of 80 W/m2 -K is provided across the chip, determine the power dissipation. Radiation 19.15: A small sphere, whose surface is an ideal emitter, has a diameter of 4 cm. It is held at 127◦ C and it radiates to a large enclosure at 27◦ C. Determine the heat transferred by radiation. 19.16: If the pipe in Problem 19.11 is an ideal emitter and it exists in a vacuum, determine the heat flow by radiation. 19.17: If the surface of the power transistor in Problem 19.13 is an ideal emitter and the enclosure surrounding the transistor is maintained at 37◦ C, determine the heat loss to the enclosure by radiation. 19.18: If the surface of the pipe in Problem 19.12 is an ideal emitter and 640 W are to be dissipated, determine the surface temperature of the pipe if radiation is the only mode of heat transfer. 19.19: If the surface of the wall in Problem 19.9 is an ideal emitter, determine the heat flow from the wall by radiation. 19.20: A steel plate that is coated so that its surface may be considered as an ideal emitter is suspended by nonconducting cords in an evacuated enclosure with walls at 27◦ C. The plate is 50-cm square. Determine the surface temperature required to make the dissipation from the plate 68.5 W. Thermal Resistance 19.21: Determine the thermal resistance of the plate in Problem 19.1. 19.22: Determine the thermal resistance of the walls of the igloo in Problem 19.3. 19.23: Determine the thermal resistance of the walls of the beaker in Problem 19.4. 19.24: Determine the thermal resistance of the walls of the beverage carrier in Problem 19.5. 19.25: Determine the thermal resistance of the pipe in Problem 19.6. 19.26: Determine the thermal resistance of the concrete floor in Problem 19.8. 19.27: Determine the convective thermal resistance of the plate in Problem 19.9. 19.28: Determine the convective thermal resistance of the sphere in Problem 19.10. 19.29: Determine the convective thermal resistance of the pipe in Problem 19.11. 19.30: Determine the radiative thermal resistance of the sphere in Problem 19.15. 19.31: The outside surface of the pipe in Problem 19.11 is an ideal emitter and it is held at 77◦ C in an enclosure at 17◦ C. Determine the radiative thermal resistance. 19.32: Determine the radiative thermal resistance of the transistor in Problem 19.17.

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Introduction to Thermal and Fluid Engineering

19.33: Determine the radiative thermal resistance of the pipe in Problem 19.18 if the pipe is held at 107◦ C and is put into an enclosure at 7◦ C. 19.34: Determine the radiative thermal resistance of the wall in Problem 19.19. Combined Mechanisms of Heat Transfer 19.35: A composite wall has an area of 48 m2 and consists of a 6.25-cm layer of fiberglass insulation (k = 0.046 W/m-K) sandwiched between a pair of 2.5-cm-thick white pine boards (k = 0.24 W/m-K). The temperatures at the face of the boards are 30◦ C and 12◦ C. Determine (a) the thermal resistance of the fiberglass, (b) the thermal resistance of each board, and (c) the heat flow. 19.36: A wall having an area of 8 m2 is fabricated of corkboard having a thickness of 1.27 cm and oak (k = 0.35 W/m-K) having a thickness of 1.905 cm. The corkboard side is in a room that is maintained at 77◦ C with a heat transfer coefficient of 5 W/m2 -K. The oak side is exposed to a winter environment at 0◦ C with a wind blowing such that the heat transfer coefficient is 12 W/m2 -K. Determine the heat transferred. 19.37: A plane wall has a surface area of 4 m2 and a surface temperature of 327◦ C. One surface of the wall is insulated and the other surface is exposed to enclosure walls and an air flow at 27◦ C with a heat transfer coefficient of 14.5 W/m2 -K. The surface may be treated as an ideal emitter. Determine the heat transferred from the wall to the surroundings. 19.38: A 2-in × 10-BWG zinc tube is 2-m long and its inside surface is held at 102◦ C. The tube is covered with a 1.75-cm layer of asbestos insulation. The asbestos is exposed to an environment at 27◦ C with a heat transfer coefficient of 40 W/m2 -K. Determine (a) the total thermal resistance, (b) the heat transferred, and (c) the temperature at the zinc-asbestos interface. 19.39: The interior of a 2.5-m long 6-in schedule 80 brass (70% copper and 30% zinc) pipe is held at 227◦ C. The pipe is covered with 1 cm of 85% magnesia insulation (k = 0.074 W/m-K) and is placed in an evacuated space at 27◦ C. The insulation is presumed to have a surface that is an ideal emitter. Determine the heat flow. 19.40: An insulated 6-in-schedule 80 pipe fabricated of 1% carbon steel and 8-m long carries a liquid at 125◦ C with a heat transfer coefficient of 242 W/m2 -K. The insulation consists of 1 cm of fiberglass (k = 0.046 W/m-K). The pipe-insulation entity is in atmospheric air at 15◦ C with a natural convection heat transfer coefficient of 14.5 W/m2 -K. Determine (a) the thermal resistance of the pipe, (b) the thermal resistance of the insulation, and (c) the heat transferred. 19.41: The interior surface of an 8-cm-outside-diameter hollow brass sphere is held at 150◦ C. The wall of the sphere is 0.75-cm thick and it is covered with a 0.625-cm layer of asbestos. The exterior of the asbestos is maintained in a 15◦ C environment where the heat transfer coefficient is 12.2 W/m2 -K. Determine (a) the heat transferred and (b) the temperature at the brass asbestos interface. 19.42: A brass sphere having a diameter of 10 cm has a surface that is considered an ideal emitter. The surface is held at 227◦ C. Convection and radiation are present at the exterior of the sphere in a 27◦ C environment (both airstream and enclosure walls) and the heat transfer coefficient is 40 W/m2 -K. Determine (a) the convection thermal resistance and (b) the heat transferred.

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607

The Overall Heat Transfer Coefficient 19.43: In Problem 19.35, what is the overall heat transfer coefficient? 19.44: In Problem 19.36, what is the overall heat transfer coefficient? 19.45: Use the surface at the exterior of the asbestos in Problem 19.38 to determine the overall heat transfer coefficient. 19.46: Use the surface at the exterior of the fiberglass in Problem 19.40 to determine the overall heat transfer coefficient. 19.47: Use the surface at the interior of the sphere in Problem 19.41 to determine the overall heat transfer coefficient. 19.48: Use the surface at the exterior of the sphere in Problem 19.42 to determine the overall heat transfer coefficient.

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20 Steady-State Conduction

Chapter Objectives •

To introduce the Fourier law and the equation that governs steady conduction in one dimension.

To analyze one-dimensional steady conduction through a plane wall, a hollow cylinder, and a hollow sphere. • To introduce contact resistance and discuss its role in conduction through composite systems. • To consider the critical radius of insulation. •

To discuss the effect of uniform heat generation on one-dimensional heat conduction. • To study the performance of longitudinal and radial fins as heat transfer enhancement devices. • To provide an introduction to two-dimensional steady conduction and to illustrate the conduction shape factor method. •

20.1

Introduction

A wide variety of engineering applications involve heat transfer by conduction. Unlike convection that pertains to energy transport due to fluid motion and radiation that can propagate in a perfect vacuum, conduction requires the presence of an intervening medium. The medium can either be a solid or a stationary fluid. At a microscopic level, conduction in stationary fluids is a consequence of molecules at higher temperature interacting and exchanging energy with molecules at a lower temperature. In solids, the transport of energy by conduction involves phenomena such as lattice waves (phonons) and translational motion of free electrons. This microscopic approach is essential in the study of the thermal behavior of superconducting thin films, microsensors, and other micromechanical devices. For a vast majority of engineering applications, a macroscopic study based on the Fourier law suffices. In the study of heat conduction, a distinction is made among uniform, nonuniform, steady, and nonsteady or transient heat flow. Usually, the words “uniform” and “nonuniform” refer to spacial variations, whereas “steady” and “transient” refer to the time domain. We note that steady-state heat flow exists when the temperature does not vary with

609

610

Introduction to Thermal and Fluid Engineering

time at any point in the heat flow path. Conversely, transient heat flow occurs when the temperatures at any point in the heat flow path varies with time.

20.2

The General Equation of Heat Conduction

It was Fourier,1 who in 1822, proposed that the rate of heat flow by conduction through a material is proportional to the area normal to the heat flow path and to the temperature gradient along the heat flow path. We assemble these two facts into a proportionality ˙ x ∼ −AdT Q dx where the minus sign is employed to allow for a positive heat flow in the presence of a falling temperature gradient along the heat flow path. Insertion of a proportionality constant gives us what is called the Fourier law ˙ x = −k AdT Q dx

(20.1)

which serves to define the proportionality constant, k, as the thermal conductivity, a property of the material k≡−

˙x Q A(−dT/dx)

With Equation 20.1 in hand, we may now derive the general equation of heat conduction by considering the differential volume shown in Figure 20.1. We write an energy balance equating the rate of heat flow entering the differential volume with the rate of heat flow leaving plus the rate of change of internal energy (actually the heat stored in the differential volume) and obtain ˙ in + Q ˙ gen = Q ˙ out + Q ˙ stored Q or

pjwstk|402064|1435600832

˙ in − Q ˙ out + Q ˙ gen = Q ˙ stored Q

(20.2)

Consider the x direction first. Heat will enter the left face in accordance with Equation 20.1   ∂T ∂T ˙ Qx = −k A =− k dydz ∂x ∂x where the temperature gradient is expressed as a partial derivative because the temperature is a function of x, y, z, and t. Heat will leave the right face at x + dx      ∂T ∂ ∂T ˙ Qx+dx = −k + −k dx dydz ∂x ∂x ∂x 1 J. B. J. Fourier (1768–1830) was a French geometrician and physicist who formalized the law of heat propagation

and proposed the Fourier series, which is a trigonometric series for arbitrary functions.

Steady-State Conduction

611 . Qy+dy

. Qz+dz dx

. Q i Generated within Volume

dy

. Qx+dx

. Qx

dz . Q Stored within Volume . Qy

. Qz

z y

x FIGURE 20.1 Differential control volume used to derive the general equation of heat conduction.

so that the difference between the heat entering at x and leaving at x + dx is   ∂ ∂T ˙ ˙ Qx − Qx+dx = k dxdydz ∂x ∂x We may use an identical procedure for the y and z directions   ∂ ∂T ˙ ˙ Q y − Q y+dy = k dxdydz ∂y ∂y and

  ˙z−Q ˙ z+dz = ∂ k ∂T dxdydz Q ∂z ∂z

˙ =Q ˙ in − Q ˙ out in Equation 20.2 is now accounted for and is Thus,  Q        ∂ ∂T ∂ ∂T ∂ ∂T ˙ Q = k + k + k dxdydz ∂x ∂x ∂y ∂y ∂z ∂z

(20.3)

The heat generated within the differential volume may be an I 2 R dissipation, or a chemical or nuclear reaction. With q i designated as the volumetric heat generation in W/m3 , ˙ gen in Equation 20.2 is Q ˙ gen = q i dxdydz Q

(20.4)

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Introduction to Thermal and Fluid Engineering

Finally, the heat stored within the differential volume is related to the rate of change of internal energy, du ∂T = c(dm) dt ∂t With dm = dxdydz, where  is the material density, we have ˙ stored = du = c ∂T dxdydz Q dt ∂t

(20.5)

When Equations 20.3, 20.4, and 20.5 are inserted into Equation 20.2, the result is       ∂ ∂T ∂ ∂T ∂ ∂T ∂T k + k + k + q i = c ∂x ∂x ∂y ∂y ∂z ∂z ∂t where the common dxdydz terms have been cancelled. Then, if k, , and c are independent of temperature, position, and time, we obtain the general equation of heat conduction ∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

+

qi 1 ∂T = k  ∂t

(20.6)

where  is called the thermal diffusivity of the material ≡

k c

(m2 /s)

(20.7)

If the system contains no heat sources, Equation 20.6 becomes the Fourier equation (not the Fourier law) ∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

=

1 ∂T  ∂t

(20.8)

If steady state is considered, that is, if the temperature distribution does not vary with time, we obtain the Poisson2 equation. ∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

+

qi =0 k

(20.9)

Finally, in the absence of heat sources and in the steady state, Equation 20.6 reduces to the Laplace3 equation ∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

=0

(20.10)

which is often written as ∇2T = 0

(20.11)

2 S. D. Poisson (1781–1840) was a French mathematician who worked on mathematical physics, contributed to the

wave theory of light, and formulated the Poisson ratio governing the elasticity of materials. 3 P.

S. Laplace (1749–1827) was a French astronomer and mathematician who contributed to celestial mechanics, formulated the laws of probability, and discovered the Laplace differential equation.

Steady-State Conduction

613

where ∇ 2 is a differential operator known as the Laplacian, which in rectangular coordinates, is ∇2 ≡

20.3

∂2

+

∂x2

∂2 ∂ y2

+

∂2 ∂z2

=0

Conduction in Plane Walls

20.3.1 The Single-Material Layer We first take the simplest possible case, shown in Figure 20.2. In the steady state where there is no heat generation and where the heat flow is in the x direction, Equation 20.11 reduces to the ordinary differential equation d2T =0 d x2

(20.12)

T = C1 x + C2

(20.13)

which can be integrated twice to yield

The specified boundary conditions are T(x = 0) = T1

T(x = L) = T2

and

T1

. Qx

T2

dx

x

x + dx L

x=0 FIGURE 20.2 Single slab used to derive the Fourier law in one dimension.

x=L

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Introduction to Thermal and Fluid Engineering

which may be applied to Equation 20.13 to obtain the pair of simultaneous linear equations T1 = C2 T2 = C1 L + C2 The solution to this set is C2 = T1

and

C1 =

T2 − T1 L

so that the temperature distribution in the plane wall is T(x) =

T2 − T1 x + T1 L

(20.14)

Two conclusions have emerged from this analysis: 1. The first integration of Equation 20.12 yields dT/dx = constant, which coupled ˙ x = constant. Thus, for steady-state one-dimensional with Equation 20.1, yields Q conduction, in a plane wall with no energy generation, the heat flux is constant. 2. The temperature distribution in a plane wall with constant thermal conductivity and no energy generation varies linearly with the thickness coordinate, x. We can evaluate dT/dx from Equation 20.14 dT T2 − T1 = dx L and use this result in Equation 20.1 to obtain ˙ x = k A(T1 − T2 ) Q L

(19.3)

Example 20.1 Consider the plane wall of Figure 20.3 with the face at x = 0 kept at T1 but with the face at x = L cooled by a fluid at temperature T∞ under a heat transfer coefficient of h. Determine the temperature distribution in the wall, the surface temperature at x = L, and the rate of heat transfer by (a) integrating the heat conduction equation and (b) using the electrothermal analog.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (a) The heat conduction equation d2T =0 d x2

(20.12)

Steady-State Conduction

615 h

k T1 . Qx

T2 T∞ x=0

x=L

FIGURE 20.3 Plane wall with convection at its right face.

is to be solved subject to the boundary conditions T(x = 0) = T1

−k

and

 dT  = h(T2 − T∞ ) dx x=L

where, for the plane wall, the cross-sectional area, A, is equal to the surface area, S. Integration of the heat conduction equation and application of the foregoing boundary conditions gives T = T1 −

h(T1 − T∞ )x ⇐ hL + k

and the temperature at the right face of the wall at x = L will be T2 = T1 −

h(T1 − T∞ )L ⇐ hL + k

˙ x. The Fourier law can now be used to determine Q ˙ x = −k AdT = k Ah(T1 − T∞ ) = T1 − T∞ ⇐ Q L 1 dx hL + k + kA hA (b) The thermal network (Figure 20.4) consists of a conduction resistance, L/k A, in a ˙ x can be series with a surface convection resistance, 1/ h S. With S = A, the value of Q . Qx

L kA T1

FIGURE 20.4 Electrothermal analog network for Example 20.1.

l hA T2

T∞

616

Introduction to Thermal and Fluid Engineering T1 L kA

T1 — T2

T2 T1 — T∞

l hA

T2 — T∞

T∞

FIGURE 20.5 Electrothermal analog network used to consider the voltage divider concept for Example 20.1.

readily obtained by dividing the overall temperature difference, T1 − T∞ , by the sum of the resistances, L/k A and 1/ h A : ˙ x = T1 − T∞ ⇐ Q L 1 + kA hA This is the rate of heat transfer established in part (a). Because ˙ x = T1 − T2 Q L kA we obtain T2 as T2 = T1 −

˙ xL Q (T1 − T∞ )L h(T1 − T∞ )L = T1 − = T1 − ⇐ kA hL + k (L/k A + 1/ h A) k A

Those who are familiar with the voltage divider concept (Figure 20.5) may note that L T1 − T2 kA = L 1 T1 − T∞ + kA hA or T2 = T1 −

h(T1 − T∞ )L hL + k

Finally, the temperature distribution can be expressed in terms of T1 , T2 , x, and L using Equation 20.14 T(x) =

T2 − T1 x + T1 L

Thus, when we employ the expression for T2 , we have, after simplification T(x) = T1 −

h(T1 − T∞ )x hL + k

Observe that the results of the two approaches are exactly the same.

Steady-State Conduction

617

Example 20.2 A glass window with a thermal conductivity of k = 1.4 W-m/K is 90-cm high, 50-cm wide, and 6-mm thick. On a winter day, forced air heating maintains the inside surface of the window at T1 (Figure 20.6). The outside surface of the window is at T2 = 10◦ C and the window loses heat by convection and radiation to the outside world at T∞ = 5◦ C. The convection heat transfer coefficient is 10 W/m2 -K and the emissivity of the outside window surface is  = 0.92. Determine (a) the convective and radiative heat losses from the outside surface and (b) the inside surface temperature of the window.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction, convection, and radiation are the modes of heat transfer. (a) In Figure 20.6, we use Equations 19.8 and 19.11 to compute the convection and radiation heat loss. In both cases, A = S = (0.90 m)(0.50 m) = 0.45 m2 . From Equation 19.8 ˙ conv = h A(T1 − T∞ ) Q = (10 W/m2 )(0.45 m2 )(10◦ C − 5◦ C) = 22.50 W ⇐ and from Equation 20.11 with T2 = 283 K and T∞ = 278 K 4 ˙ rad = A(T14 − T∞ Q )

= (0.92)(5.67 × 10−8 W/m2 -K4 )(0.45 m2 )[(283 K) 4 − (278 K) 4 ] = 10.36 W ⇐ (b) An energy balance at the outer surface provides ˙ cond = Q ˙ conv + Q ˙ rad Q h

pjwstk|402064|1435600837

T1

. Qconv . Qrad T2 T∞

FIGURE 20.6 Configuration for Example 20.2.

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Introduction to Thermal and Fluid Engineering

or k A(T1 − T2 ) = 22.50 W + 10.36 W = 32.86 W L We may then solve for T1 T1 = (32.86 W) =

L + T2 kA

(32.86 W)(0.006 m) + 10◦ C = 10.31◦ C ⇐ (1.4W/m-)(0.45m2 )

20.3.2 The Composite Plane Wall Most engineering applications involve heat conduction through composite walls consisting of two or more layers of material. A typical example is the building wall, which is a composite of plaster board, fiberglass insulation and either brick, wood, or aluminum siding. Figure 20.7 shows a composite multilayered wall consisting of n layers and with heat flow from left to right. The ith layer has thickness, L i and thermal conductivity, ki , and the temperatures at its extremes are T1 and Tn+1 . We allow for a convective heat input at the left (from T∞,1 with h 1 ) and a heat loss at the right (to T∞,n+1 with h n+1 ). The electrothermal analog for the n-layer wall is shown in Figure 20.8a. The heat flow is clearly the same in all layers (convective and conductive) and the total thermal resistance of the n-conduction layers and two convection layers will be RT =

n 1 Li 1 + + h 1 S i=1 ki A h n+1 S

hn+1

h1 k1

k2

kn

T∞,1 T1 T2 . Qx T3

Tn Tn+1

L1

L2

FIGURE 20.7 Composite plane wall with convection on both exterior faces.

Ln

T∞,n+1

Steady-State Conduction

619 T∞,1 R=

1 h1A

T1 R=

L1 k1A

T∞,1

L2 k2A

T1

1 h 1A

T2 . Qx

R=

L1 k1A

T3 . Qx

Tn R=

T2

Ln knA

L2 k2A T3

Tn+1 R=

1 h 3A

1 hn+1A

T∞,n+1

T∞,3

(a)

(b)

FIGURE 20.8 Electrothermal analog network with convection for (a) the n-layer wall and (b) the two-layer wall.

˙ x , will be T/RT Then, with S = A, we find that the heat flow, Q ˙x= Q

T∞,1 − T∞,n+1 n 1 Li 1 + + h 1 S i=1 ki A h n+1 S

(20.15)

and if we have just a pair of plane layers, again, with convection, (Figure 20.9) with the electrothermal analog in Figure 20.8b, ˙ x = T∞,1 − T∞,n+1 = Q RT

T∞,1 − T∞,n+3 1 L1 L2 1 + + + h 1 A k1 A k2 A h 3 A

Sometimes, the composite wall construction is such that its electrothermal analog network involves resistances in parallel as well as in a series. Such a scenario is illustrated in Figure 20.10. If materials 2 and 3 have comparable thermal conductivities, the heat conduction may be assumed to be one dimensional, that is, the surfaces normal to the direction of heat conduction are isothermal. With the height of each component designated as H and with Z taken as the width normal to the plane of the paper, the rate of heat transfer by conduction can be expressed as ˙x Q = W

T1 − T4 L1 L2 L3 L4 + + k1 H1 k2 H2 L 3 + k3 H3 L 2 k4 H4

(20.16a)

620

Introduction to Thermal and Fluid Engineering h3

h1

k1

T∞,1

k2

T1 . Qx T2 T3 L1

T∞,3

L2

FIGURE 20.9 The composite wall of two layers with convection at both exterior surfaces.

or because L 2 = L 3 ˙x Q = Z

T1 − T4 L1 L2 L4 + + k1 H1 k2 H2 + k3 H3 k4 H4

(20.16b)

where H1 = H2 + H3 = H4 and the rule for combining resistances in parallel R =

R1 R2 R1 + R2

has been employed. L2 1

2

4

H2 T4

T1 H1

H4 3 H3

L1

L3

FIGURE 20.10 Multicomponent plane wall with heat flow paths in parallel.

L4

Steady-State Conduction

621 R2 = R1 =

. Qx

L2 k2H2Z

L1 k1H1Z

R4 =

L4 k4H4Z

T1

T4

R3 =

L3 k3H3Z

FIGURE 20.11 Electrothermal analog for composite wall in Example 20.3.

˙ x has been determined, the interface temperatures, T1 and T2 , may be established Once Q  ˙   Qx L1 W k1 H1

(20.17)

 ˙   Qx L4 T3 = T4 + W k4 H4

(20.18)

T2 = T1 − and

The foregoing is illustrated in Example 20.3 that now follows.

Example 20.3 Determine the rate of conduction heat flow through the composite wall

structure shown in Figure 20.10. The wall dimensions are L 1 = 5 cm, H1 = 12 cm, L 2 = L 3 = 10 cm, H2 = 5 cm, H3 = 7 cm, L 4 = 4 cm, and W = 1 m. The thermal conductivities may be taken as k1 = 2 W/m-K, k2 = 0.4 W/m-K, k3 = 0.5 W/m-K, and k4 = 12 W/m-K and the surface temperatures are T1 = 80◦ C and T4 = 30◦ C (Figure 20.10).

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction is the only mode of heat transfer. (5) There is no contact resistance between any of the components. The thermal resistances are calculated first: R1 =

L1 0.05 m = = 0.2083 K/W k1 H1 W (2 W/m-K)(0.12 m)(1 m)

R2 =

L2 0.10 m = = 5.0000 K/W k2 H2 W (0.4 W/m-K)(0.05 m)(1 m)

R3 =

L3 0.10 m = = 2.8571 K/W k3 H3 W (0.5 W/m-K)(0.07 m)(1 m)

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Introduction to Thermal and Fluid Engineering

R1

Req

T1 – T4

R4

FIGURE 20.12 Simplification of electrothermal analog network for composite wall in Example 20.3.

and R4 =

L4 0.04 m = = 0.0278 K/W k4 H4 Z (12 W/m-K)(0.12 m)(1 m)

The equivalent of R2 and R3 in parallel will be Req (Figures 20.11 and 20.12) Req =

R2 R3 (5.0000 K/W)(2.8571 K/W) = = 1.8182 K/W R2 + R3 5.0000 K/W + 2.8571 K/W

Then, use of the electrothermal analog network in Figure 20.12 provides the rate of heat transfer ˙x= Q

T1 − T4 80◦ C − 30◦ C = = 24.3 W ⇐ R1 + Re + R4 (0.2083 + 1.8182 + 0.0278) K/W

20.3.3 Contact Resistance The conduction analysis for a composite wall presented in Section 20.3.2 assumed a perfect contact at the interface between the two materials. In reality, the mating surfaces are rough and the actual contact occurs at discrete points (asperities or peaks) as illustrated in Figure 20.13. The voids between the two surfaces are filled with air or some other material. As indicated in Figure 20.14, the departure from perfect contact is manifested in the form of a temperature drop, Tc in K, across the contact and the ratio of this temperature drop

Intimate Contact Gap Filled with Fluid with Thermal Conductivity kf

FIGURE 20.13 Artist’s conception of two metals in contact.

Steady-State Conduction

623 Line of Ideal Contact

T

. qx

ΔTc

x FIGURE 20.14 Temperature drop across contacting surfaces.

to the heat flux, q˙ x in W/m2 , is defined as the contact resistance, Rc =

Tc q˙ x

(K-m2 /W)

(20.19)

Because Rc is defined on the basis of a heat flux and not the rate of heat transfer, it should be divided by the contact area, A, and then incorporated into the construction of the electrothermal network. The parameter Rc is a complicated function of the contact pressure, surface roughnesses, thermal conductivities, the microhardness of the mating surfaces and the material trapped in the voids. Additional complications arise when bolts, screws, or rivets are used to fasten the surfaces. An excellent review is provided by Yovanovich and Marotta in Bejan and Kraus (2001). In the thermal control of microelectronic equipment, it is crucial to minimize the contact resistance between mating surfaces. This is achieved by treating the surfaces with siliconbased thermal grease or a soft metallic foil prior to the fastening of the two materials. Table 20.1 provides some typical contact resistances of metal joints.

Example 20.4 A composite wall is made of two materials with thermal conductivities

k1 = 5 W/m-K and k2 = 7 W/m-K and with thicknesses L 1 = 15 mm and L 2 = 25 mm TABLE 20.1

Contact Resistance of Metal Joints in the Air at Moderate Temperatures Roughness, m

Contact Pressure, MPa

Rc , K-m2 /W

Ground 416 stainless steel Ground 304 stainless steel Ground aluminum Ground copper Ground 416 stainless steel Stainless steel-aluminum

2.54 1.14 2.54 1.27 2.54 20–30

Stainless steel-aluminum

1.3–2.0

Ground aluminum-copper

1.3–1.4

Milled aluminum-copper

4.4–4.5 4.4–4.5

0.3–2.5 4–7 1.2–2.5 1.2–20 0.30–2.50 10 20 10 20 5 15 10 20–35

2.63 × 10−4 5.26 × 10−4 8.77 × 10−5 6.99 × 10−6 2.63 × 10−4 3.45 × 10−4 2.78 × 10−4 6.10 × 10−5 4.81 × 10−5 2.38 × 10−5 1.79 × 10−5 8.33 × 10−5 4.45 × 10−4

Material

624

Introduction to Thermal and Fluid Engineering k1 = 5 W/m-K

k2= 7 W/m-K

T∞,1 T1

h3 = 25 W/m2-K h1 = 50 w/m2-K

ΔTc

T3 T∞,3 L1 = 15 mm

L2 = 25 mm

FIGURE 20.15 Configuration for Example 20.4.

(Figure 20.15). The left face of the wall is exposed to a hot fluid at T∞,1 = 80◦ C with a heat transfer coefficient of h 1 = 50 W/m2 -K. The right face is in contact with cold air at T∞,3 = 25◦ C with a heat transfer coefficient of h 3 = 25 W/m2 -K. The wall area normal to the heat flow direction is A = 0.80 m2 . Determine the rate of heat transfer, the surface temperatures, T1 and T3 , and the temperature drop, Tc , across the contact assuming (a) perfect contact and (b) allowing for a contact resistance of Rc = 0.02 m2 -K/W.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (a) Equation 20.15 is applicable, and for the case of no contact resistance may be used to ˙ x. compute Q ˙x= Q

T∞,1 − T∞,3 1 L1 L2 1 + + + h 1 A k1 A k2 A h 3 A

The individual resistances in the denominator are 1 1 = = 0.0250 K/W 2 h1 A (50 W/m -K)(0.80 m2 ) L1 0.015 m = = 3.750 × 10−3 K/W k1 A (5 W/m-K)(0.80 m2 ) L2 0.025 m = = 4.464 × 10−3 K/W k2 A (7 W/m-K)(0.80 m2 )

Steady-State Conduction

625

and 1 1 = = 0.0500 K/W 2 h 3 A (25 W/m -K)(0.80 m2 ) With these values put into Equation 20.15, we obtain ˙x = Q =

80◦ C − 25◦ C (25.000 + 3.750 + 4.464 + 50.000) × 10−3 K/W 55◦ C = 661 W ⇐ 0.0832 K/W

The temperatures, T1 and T2 , derive from Ohm’s law in a thermal sense T1 = T∞,1 −

˙x Q 661W = 80◦ C − = 63.5◦ C ⇐ h1 A (50 W/m2 -K)(0.80 m2 )

T3 = T∞,3 +

˙x Q 661 W = 25◦ C + = 58.1◦ C ⇐ h2 A (25 W/m2 -K)(0.80 m2 )

and

and for a perfect contact Tc = 0 ⇐ (b) With the inclusion of contact resistance, Equation 20.15 must be modified to ˙x= Q

T∞,1 − T∞,2 1 L1 Rc L2 1 + + + + h 1 A k1 A A k2 A h 3 A

With Rc 0.02 m2 -K/W = = 0.0250 K/W A 0.80 m2 we have ˙x= Q

80◦ C − 25◦ C 55◦ C = = 508 W ⇐ (0.0832 K/W + 0.0250 K/W) 0.1082 K/W

pjwstk|402064|1435600849

and we note that the presence of the contact resistance reduces the rate of heat transfer by about 23%. We again find T1 and T2 from Equation 20.21 T1 = T∞,1 −

˙x Q 508 W = 80◦ C − = 67.3◦ C ⇐ h1 A (50 W/m2 -K)(0.80 m2 )

T3 = T∞,3 +

˙x Q 508 W = 25◦ C + = 50.4◦ C ⇐ h3 A (25 W/m2 -K)(0.80 m2 )

and

Finally, Equation 20.19 permits us to find Tc Tc =

˙x Rc Q (0.02m2 -K/W)(508 W) = 12.7◦ C ⇐ = A 0.80 m2

626

20.4

Introduction to Thermal and Fluid Engineering

Radial Heat Flow

20.4.1 Cylindrical Coordinates The Laplace equation in cylindrical coordinates can be developed in a manner similar to the development in Section 20.2. In the single, radial dimension, this equation becomes the ordinary differential equation   d dT r =0 (20.20) dr dr and we use this equation to develop the heat flow in the radial direction and the temperature distribution in a variety of cylindrical configurations. 20.4.1.1 The Hollow Cylinder Figure 20.16 shows the end view of a hollow cylinder of inside radius, r1 , outside radius, r2 , length, L, and thermal conductivity, k. The inside and outside surfaces are maintained at the constant temperatures T1 and T2 and the heat flows in the radial direction. As indicated in Section 20.4.1, the differential equation that applies in this case is Equation 20.20. The boundary conditions are T(r = r1 ) = T1

and

T(r = r2 ) = T2

and solution of this equation with the foregoing boundary conditions yields ˙ r = 2k L(T1 − T2 ) Q ln(r2 /r1 )

(20.21)

Here it is worth noting that the temperature distribution is logarithmic and that the heat flux, q˙ r , varies inversely with the radius but the rate of heat transfer is independent of the radius.

r + dr r dr

k T2

T1 L

r1 r2 FIGURE 20.16 Hollow cylinder used to establish the Fourier law in cylindrical coordinates.

Steady-State Conduction

627 T∞,1 = 45.81°C h = 3000 W/m2-K

1 mm

T∞,2 = 20°C

1.25 cm

h = 250 W/m2-k

k = 111 W/m-k FIGURE 20.17 Hollow cylinder for Example 20.5.

The thermal resistance for the wall of a hollow cylinder is identified by rewriting Equation 20.21 and obtaining Equation 19.12b Rcond =

T1 − T2 ln(r2 /r1 ) = ˙ 2k L Qr

(20.22)

and if convection occurs on the inside of the cylinder (at T∞,1 with h 1 ) and on the outside of the cylinder (at T∞,2 with h 2 ), then the rate of heat transfer may be expressed as ˙r = Q

T∞,1 − T∞,2 1 ln(r2 /r1 ) 1 + + h 1 (2r1 L) 2k L h 2 (2r2 L)

(20.23)

Example 20.5 A typical condenser in a steam power plant consists of hundreds (and sometimes thousands) of metal tubes that are housed in a shell. Steam condenses on the outside of the tubes as the cooling water flows inside. Figure 20.17 shows one such tube, made of brass (k = 111 W/m-K) and having an inside diameter of 2.5 cm and a thickness of 1 mm. The temperature of the condensing steam is 45.81◦ C and the associated heat transfer coefficient is 3000 W/m2 -K. The cooling water, at an average temperature of 20◦ C, provides a convection heat transfer coefficient of 250 W/m2 -K. Determine (a) the rate of heat transfer from the steam to the cooling water per meter of length of tube and (b) the rate of condensation of the steam per meter of length of tube if h fg = 2392.8 kJ/kg.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer.

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Introduction to Thermal and Fluid Engineering

Here d1 2.500 cm = = 1.250 cm and r2 = r1 + 0.100 cm = 1.350 cm 2 2

r1 =

(a) Equation 20.23 with T∞,2 and T∞,1 reversed because the heat flow is inward ˙r = Q

T∞,2 − T∞,1 1 ln(r2 /r1 ) 1 + + h 1 (2r1 L) 2k L h 2 (2r2 L)

(20.23)

˙ r . With L = 1 m, we can compute the individual thermal may be used to determine Q resistances as 1 1 = = 0.0509 K/W 2 h 1 (2r1 L) (250 W/m -K)(2)(0.0125 m)(1 m) ln(r2 /r1 ) ln(0.0135/0.0125) = = 1.103 × 10−4 K/W 2k1 L (2)(111 W/m-K)(1 m) and 1 1 = = 3.930 × 10−3 K/W h 2 (2r2 L) (3000 W/m2 -K)(2)(0.0135 m)(1 m) Then, we have ˙r = Q

45.81◦ C − 20◦ C 0.0509 K/W + 1.103 × 10−4 K/W + 3.930 × 10−3 K/W

and with a total resistance of 0.0550 K/W, the result is ˙ r = 469.5 W ⇐ Q (b) The rate of condensate per meter of length of tube is m ˙ =

˙r Q 469.5 W = = 1.962 × 10−4 kg/s ⇐ h fg 2392.8 kJ/kg

20.4.1.2 The Composite Hollow Cylinder Figure 20.18a shows a composite hollow cylinder composed of n layers. The composite cylinder is heated by convection on the inside surface at r1 where the heat transfer coefficient is h 1 and the temperature of the fluid on the inside is T∞,1 . The outer cylinder is convectively cooled at rn+1 , where the heat transfer coefficient is h n+1 and the temperature of the fluid on the outside is T∞, n+1 . Heat flow occurs in the radially outward direction. The electrothermal analog network of Figure 20.18b facilitates the analysis. The rate of heat transfer for n layers can be expressed in the form of Equation 20.23 ˙r = Q

T∞,1 − T∞,n+1

n 1 ln(ri+1 /ri ) 1 + + h 1 (2r1 L) 2ki L h n+1 (2rn+1 L) i=1

(20.24)

Example 20.6 Figure 20.19 shows a steel steam pipe (k = 15 W/m-K) with an inside radius of 10 cm and an outside radius of 12 cm covered with a 3-cm-thick layer of calcium silicate with a thermal conductivity of k = 0.2 W/m-K. A 2-cm layer of fiberglass

Steady-State Conduction

629

T∞,n+1 h1 k1 k2 r1

kn

r2

r3

rn

rn+1

hn+1

T∞,1 (a) T∞,1 1 h1(2πr1L) T1 ln(r2/r1) 2πk1L T2

T∞,1–T∞,n+1

ln(r3/r2) 2πk2L

+ –

T3 Tn ln(rn+1/rn) 2πknL Tn+1 1 hn+1(2πrn+1L) T∞,n+1 (b)

FIGURE 20.18 (a) Composite n-layer hollow cylinder with convection on both the interior and exterior surfaces and (b) electrothermal analog network.

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Introduction to Thermal and Fluid Engineering

Steel Pipe Calcium Silicate Sheet Fiberglass

T4 k1

h1

T∞,4

h4

k3 k2

T3

T1

T2

T∞,1

r1 r2 r3 r4 FIGURE 20.19 Three-layer composite hollow cylinder with convection for Example 20.6.

(k = 0.06 W/cm-K) covers the calcium silicate. The steam on the inside, which is at 500◦ C, provides a heat transfer coefficient of 1500 W/m2 -K and the outside surface of the fiberglass is cooled by natural convection with a heat transfer coefficient of 10 W/m2 -K. The outside surrounding temperature is 20◦ C. Determine (a) the heat loss per meter of length and (b) the temperature on the outside surface of the fiberglass.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (5) There is no contact resistance between any of the components. The configuration is shown in Figure 20.19 and we observe that because there are three “solid” layers, we apply Equation 20.23. The five individual resistances can be evaluated as 1 1 = = 1.061 × 10−3 K/W 2 h 1 (2r1 L) (1500 W/m -K)(2)(0.100 m)(1 m) ln(r2 /r1 ) ln(0.12/0.10) = = 1.934 × 10−3 K/W 2k1 L (2)(15 W/m-K)(1 m) ln(r3 /r2 ) ln(0.15/0.12) = = 0.1776 K/W 2k2 L (2)(0.2 W/m-K)(1 m) ln(r4 /r3 ) ln(0.17/0.15) = = 0.3320 K/W 2k3 L (2)(0.06 W/m-K)(1 m)

Steady-State Conduction

631

and 1 1 = = 0.0936 K/W 2 h 4 (2r4 L) (10 W/m -K)(2)(0.17 m)(1.00 m) Thus, RT = (1.061 × 10−3 K/W + 1.934 × 10−4 K/W + 0.1776 K/W + 0.3320K/W + 0.0936 K/W) = 0.6062 K/W so that ◦ ◦ 480◦ C ˙ r = 500 C − 20 C = Q = 792 W ⇐ RT 0.6062 K/W

(b) T4 will be T4 = T∞,4 +

˙r Q = 20◦ C + (792 W)(0.0936 K/W) = 94.1◦ C ⇐ 2h 4r4 L

20.4.1.3 The Critical Radius of Insulation We now consider Figure 20.20a, which shows a cylinder with a layer of insulation. Figure 20.20b provides the electrothermal analog network for just the insulation of radius, r2 , which completely surrounds the cylindrical pipe or tube of outer radius, r1 . The thermal conductivity of the insulation is k and the heat transfer coefficient at r2 is h. We observe from Figure 20.20b that the total resistance to the flow of heat will be R=

ln(r2 /r1 ) 1 + 2k L 2r2 h L

and because ˙ r = T1 − T2 Q R we see that the heat flow will be at a maximum when R is at a minimum or when   dR d ln(r2 /r1 ) 1 = + =0 dr2 dr2 2k L 2r2 h L The differentiation is straightforward and the result is that the minimum resistance, known as the critical radius of insulation for the cylinder, rcr,cyl , is rcr,cyl =

k h

(20.25)

and it can be shown that for a sphere rcr,sphere =

2k h

(20.26)

Thus, for enhancing the heat transfer, we make r < rcr,cyl

or r
rcr,cyl

or r >

k h

20.4.2 Spherical Coordinates The Laplace equation in spherical coordinates can be developed in a manner similar to the development in Section 20.2. In the single, radial dimension, this equation becomes the ordinary differential equation   d dT r2 =0 dr dr

(20.27)

and we use this equation to develop the heat flow in the radial direction and the temperature distribution in a variety of spherical configurations.

Steady-State Conduction

633 k T2

T1

r1 r2 FIGURE 20.21 Hollow sphere.

20.4.2.1 The Hollow Sphere Figure 20.21 shows a hollow sphere of inside radius, r1 , outside radius, r2 , and thermal conductivity, k. The inside and outside surfaces are maintained at the constant temperatures T1 and T2 and the heat flows in the radial direction. The differential equation that applies in this case is Equation 20.27 with the boundary conditions T(r = r1 ) = T1

and

T(r = r2 ) = T2

Solution of Equation 20.27 with the foregoing boundary conditions yields T = T1 +

T1 − T2 1 1 − r2 r1



1 1 − r1 r

 ⇐

(20.28)

and the heat flow will be ˙ r = 4k(T1 − T2 ) Q 1 1 + r1 r2

(20.29)

Equation 20.28 shows that the temperature distribution is hyperbolic, that is, it varies inversely with r. The thermal resistance of a hollow sphere can be identified from Equation 20.29 (or Equation 19.12c) as

Rcond

1 1 − r1 r2 = 4k

(20.30)

and if convection occurs on the inside of the sphere from T∞,1 via h 1 and at the outside of the sphere to T∞,2 via h 2 , then the rate of heat transfer may be expressed with 1 1 r2 − r1 − = r1 r2 r1 r2

634

Introduction to Thermal and Fluid Engineering T∞,n+1 h1 k1 k2 r1

kn

r2

r3

rn+1

rn

hn+1

T∞,1

FIGURE 20.22 Composite hollow sphere with n layers and interior and exterior convection.

as ˙r = Q

T∞,1 −, T∞,2 1 r2 − r1 1 + + 4kr1r2 4h 1r12 4h 2r22

(20.31)

20.4.2.2 The Composite Hollow Sphere The composite n-layer hollow sphere with n layers shown in Figure 20.22 (identical to Figure 20.20) is heated by convection on the inside surface and cooled by convection on the outside surface. Heat flow occurs in the radially outward direction. The fluid inside the sphere at r1 is at T∞,1 , the inner surface is at T1 , and the heat transfer coefficient at the inside of the inner shell is h 1 . The outside surface is at rn+1 where the temperature is Tn+1 and where the heat transfer coefficient is h n+1 . The surrounding temperature on the outside of the composite sphere is at T∞,n+1 . To facilitate the analysis, Figure 20.23 provides the electrothermal analog network. With

pjwstk|402064|1435600855

1 1 ri+1 − ri − = ri ri+1 ri ri+1

T∞,1

1 4πh1r12

T1

r2 — r1 4πk1r1r2

T2

r3 — r2 4πk2r3r2

T3

Tn

rn+1 — rn 4πknrn+1rn

Tn+1

1 4πhn+1rn+12 T∞,n+1

h1 + –

R1

R2

Rn+1

h2

T∞,1 — T∞,n+1

FIGURE 20.23 Electrothermal analog for the n-layer composite hollow sphere with both interior and exterior convection.

Steady-State Conduction

635 Unknown h T∞,2 = 20°C

r1 = 25 cm

Thin Wall

r2 = 30 cm

N2 at T∞,1 = –196°C

h = 0.02 w/m-k

. Q

T∞,n+1

FIGURE 20.24 Configuration for Example 20.7.

the rate of heat transfer can written as ˙r = Q

T∞,1 − T∞,n+1 n 1 ri+1 − ri 1 + + 2 4k r r 4h 1r12 4h i i i+1 n+1 r n+1 i=1

(20.32)

Example 20.7 Liquid nitrogen at −196◦ C is contained in a thin spherical tank with an outside radius of 25 cm (Figure 20.24). The tank is covered with a 5-cm-thick layer of insulation having a thermal conductivity of k = 0.02 W/m-K. The heat gain from the environment causes the liquid nitrogen to vaporize and the nitrogen vapor escapes from the tank through a vent. Determine the maximum natural convection heat transfer coefficient that can be permitted on the outside surface of the insulation if the vapor loss is to be limited to 4 × 10−4 kg/s. Take the surrounding air temperature as 20◦ C with the boiling point of nitrogen as −196◦ C and its enthalpy of vaporization as 198 kJ/kg.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (5) There is no contact resistance between any of the components. ˙ can be found from the rate of vaporization of The rate of heat transfer into the tank, Q, the nitrogen, m ˙ and the enthalpy of vaporization, h fg . Thus, ˙ = mh Q ˙ fg = (4 × 10−4 kg/s)(198 × 103 J/kg) = 79.2 W We assume that the inside surface of the spherical tank is at the temperature of the boiling nitrogen and that the temperature drop through the tank wall is negligible. With the heat

636

Introduction to Thermal and Fluid Engineering

flowing radially inward, we may therefore modify Equation 20.32 with r1 = 0.25 m and r2 = 0.30 m to ˙r = Q

T∞,2 − T∞,1 r2 − r1 1 + 4k2r2r1 4hr32

The terms in the denominator can be evaluated as r2 − r1 0.30 m − 0.25 m = = 2.6526 K/W 4k2r2r3 4(0.02 W/m-K)(0.30 m)(0.25 m) and 1 1 0.8842 = = K/W 2 2 4h(0.30 m) h 4hr2 Thus, the heat flow equation can be written as 79.2 W =

20◦ C − (−196◦ C) 0.8842 2.6526 K/W + K/W h

and we find that h = 11.84 W/m2 -K ⇐ Because radiation and convection have comparable effects in a natural convection application, the maximum limit on the natural convection heat transfer coefficient should be conservatively put as 50% of the calculated value of h. Thus, h should not exceed about 5.92 W/m2 -K if the boil-off rate is not to exceed 4 × 10−4 kg/s. Coating the surface of the insulation with a low-emissivity material would reduce the radiative heating.

20.5

Simple Shapes with Heat Generation

20.5.1 The Plane Wall The analysis for the plane wall considered in Section 20.3.1 is now extended to include the effect of internal heat generation. Such heat generation may be due to the passage of electric current or due to chemical or nuclear activity in the wall. Suppose the plane wall of Figure 20.25 is subjected to a volumetric heat generation at the rate of q i in W/m3 . In this case, Equation 20.9, which is the Poisson equation, pertains. When this relation is written in one dimension, an ordinary differential equation results d2T qi + =0 d x2 k

(20.33)

and the solution of this equation is subject to the boundary conditions T(x = −L) = T1

and

T(x = L) = T2

Steady-State Conduction

637

qi

T1 T2

x = —L

x=0

x=L

FIGURE 20.25 Plane wall with volumetric heat generation, q i .

The solution to this equation is obtained via a double integration as T=

qi 2 T2 − T1 x T1 + T2 (L − x 2 ) + + 2k 2 L 2

(20.34)

Observe that T is a quadratic function of x and that the heat flux at any location will be   dT qi x T2 − T1 q˙ = −k = −k − + dx k 2L or q˙ = q i x + k

T1 − T2 2L

(20.35)

Equation 20.35 clearly shows that q˙ is location dependent. By setting dT/dx = 0 in Equation 20.34, we can obtain the location, xmax , at which the maximum temperature occurs xmax =

k(T2 − T1 ) 2q i L

(20.36)

which when used in Equation 20.34 yields an expression for the maximum temperature in the wall: Tmax =

qi L 2 k(T2 − T1 ) 2 T1 + T2 + + 2k 8q i L 2 2

(20.37)

If the surface temperatures are identical, that is, if T1 = T2 = Ts , then the temperature distribution is symmetrical about x = 0 and it follows from Equation 20.34 that T(x) =

qi 2 (L − x 2 ) + Ts 2k

(20.38)

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Introduction to Thermal and Fluid Engineering

and the maximum temperature at the mid-plane where x = 0 will be Tmax =

qi L 2 + Ts 2k

(20.39)

Moreover, if identical convection conditions prevail on both faces so that T∞,1 = T∞,2 = T∞ and h 1 = h 2 = h, then it follows that Ts = T∞ +

qi L h

(20.40)

In this case, the temperature distribution provided by Equation 20.38 may be expressed as

x 2 T(x) − To = Ts − To L where To is the temperature at x = 0. A practical situation is one for which the temperature of the adjoining fluid, T∞ , is known. In this case, heat flows between the wall at Ts and the fluid at T∞ through a convective layer represented by h. In this case, the energy balance reduces to −k

dT |x=L = h(Ts − T∞ ) dx

and use of Equation 20.38 to establish the temperature gradient at x = L gives Ts = T∞ +

qi L h

(20.41)

Example 20.8 Rectangular copper bus bars are commonly used in the transmission of electric power. Suppose that such a bar is 10-mm thick, 200-mm high, and 3-m deep as shown in Figure 20.26. The bus bar, which carries a current of 5000 A, is exposed on its 3-m × 200-mm faces to a convective environment at 20◦ C with a heat transfer coefficient of 25 W/m2 -K. The electrical resistivity and the thermal conductivity of the copper are e = 1.68 × 10−8 -m and k = 400 W/m-K, respectively. Determine (a) the volumetric rate of heat generation, (b) the surface temperature, Ts , and (c) the maximum temperature, Tmax .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (a) The electrical resistance of the bus bar, R, may be calculated from a knowledge of its resistivity, e , depth, D, and cross-sectional area, A R = e

D (3 m) = (1.68 × 10−8 -m) = 2.52 × 10−5  A (0.20 m)(0.01 m)

The volumetric heat generation rate, q i , will then be qi =

I2R (5000 A) 2 (2.52 × 10−5 ) = = 105,000 W/m3 ⇐ volume (3 m)(0.01 m)(0.20 m)

Steady-State Conduction

639

3m

Tmax TS 200 mm

TS h, T∞ 10 mm

FIGURE 20.26 Configuration for Example 20.8.

(b) The surface temperature can be obtained from Equation 20.41 with L = 0.01 m/2 = 0.005 m. Ts = T∞ +

qi L (105,000 W/m3 )(0.005 m) = 20◦ C + = 41◦ C ⇐ k (25 W/m2 -K)

(c) The maximum temperature derives from Equation 20.39 Tmax =

qi L 2 (105,000 W/m3 )(0.005 m) 2 + Ts = + 41◦ C ≈ 41◦ C ⇐ 2k 2(400 W/m-K)

Observe that because of the high-thermal conductivity and small thickness of the bus bar, there is but a negligible difference between the maximum temperature and the surface temperature.

20.5.2 The Cylinder Suppose that the cylindrical shell of Figure 20.27 is subjected to a volumetric heat generation of q i (W/m3 ). In this case, the Poisson equation in cylindrical coordinates in one dimension gives the governing differential equation   1d dT qi r + =0 r dr dr k

(20.42)

640

Introduction to Thermal and Fluid Engineering T2

qi

k r2 r1

T1

FIGURE 20.27 Hollow cylinder with volumetric heat generation, q i .

Equation 20.42 may be solved with the boundary conditions T(r = r1 ) = T1

and

T(r = r2 ) = T2

to give qi T(r ) = T1 + 2k



r22 − r12 r + r12 ln 2 r2

 (20.43)

Next, consider a long solid cylinder of radius, ro , generating q i (W/m3 ) and whose surface is maintained at temperature, Ts . In this case, Equation 20.42 must be solved subject to the boundary conditions  dT  T(r = ro ) = Ts and =0 dr r =0 where dT/dr |r =0 is a consequence of thermal symmetry about r = 0. The solution for T takes the form T = Ts +

qi 2 (r − r 2 ) 4k o

(20.44)

and the maximum temperature occurs along the centerline (at r = 0) Tmax = Ts +

q i ro2 4k

(20.45)

If the outside surface of the cylinder is cooled by a coolant at temperature, T∞ , that provides a heat transfer coefficient, h, then, after simplification, the overall energy balance equating the heat generated, ro2 Lq i , to the heat loss by convection, (2o L)h(Ts − T∞ ), gives Ts = T∞ +

q i ro 2h

(20.46)

20.5.3 The Sphere For the spherical shell of Figure 20.28 subjected to a volumetric heat generation of q i (W/m3 ), the Poisson equation in spherical coordinates in one dimension gives the

Steady-State Conduction

641 T2

qi

k r2 r1

T1

FIGURE 20.28 Hollow sphere with volumetric heat generation, q i .

governing differential equation   d dT qi r 2 r2 + =0 dr dr k

(20.47)

When this equation is solved with the boundary conditions T(r = r1 ) = T1 the result is

and

T(r = r2 ) = T2

 2   q i r13 1 qi r 1 T = T2 + 1− + − 6k r2 3k r2 r

(20.48)

Next, consider a solid sphere of radius, ro , with surface temperature, Ts , and having an internal heat generation of q i (W/m3 ). In this case, Equation 20.47 pertains but the boundary conditions are  dT  T(r = ro ) = Ts and =0 (20.49) dr r =ro The solution for T takes the form T = Ts +

qi 2 (r − r 2 ) 6k o

(20.50)

and the maximum temperature occurs at the center where r = 0 Tmax = Ts +

q i ro2 6k

(20.51)

If a coolant at temperature T∞ provides a heat transfer coefficient of h at the outside surface of the sphere, then we can equate the total rate of heat generation to the rate of convection from the outside surface at r = ro   4 3 qi r = h(4ro2 )(Ts − T∞ ) 3 o or Ts = T∞ +

q i ro 3h

(20.52)

642

20.6

Introduction to Thermal and Fluid Engineering

Extended Surfaces

20.6.1 Introduction When engineers need to increase the heat transfer between a primary surface and its adjoining fluid, it is a common practice to attach an extended surface or fin to the primary surface. Although the extended surface provides an increased heat transfer surface area, it also decreases the average primary surface temperature. However, if the extended surface is properly designed, the net result is an increase in the rate of heat transfer. Extended surfaces appear in such diverse applications as air-cooled engines, compact heat exchangers, heat sinks for electronic equipment cooling, and space radiators. Decades of research in extended surface technology has resulted in a wide variety of designs. The three basic extended surface elements that appear in most designs are the longitudinal fin of uniform thickness or rectangular profile (Figure 20.29a), the cylindrical spine or pin fin (Figure 20.29b), and the annular or radial fin of uniform thickness (Figure 20.29c). We study these three configurations in this section, both as individual fins and arrays of fins. We will analyze these fins using the limiting assumptions proposed by Murray (1938) and Gardner (1945) which involve: •

A homogeneous fin material and operation in the steady state.

•

Uniform surrounding temperature.

•

Uniform heat transfer coefficient and uniform thermal conductivity.

•

No temperature gradients along the length or across the thickness of the fin.

•

No bond resistance between the primary surface and the fin.

•

No heat sources within the fin itself.

•

No heat transfer from or to the fin tip.

•

No heat transfer from the fin edges.

•

Uniform temperature along the fin base.

20.6.2 The Longitudinal Fin of Uniform Thickness Consider the longitudinal fin of thickness, , height, b, and length, L as shown in Figure 20.30. The fin is made of a material having a thermal conductivity, k. The adjoining fluid at temperature, T∞ , provides a convective heat transfer coefficient, h, over the exposed surface of the fin. To derive the differential equation that governs the temperature distribution in the fin, we focus on a differential element of height, dx, and write an energy balance giving ˙ ˙ x − h P(T − T∞ )dx = Q ˙ x+dx = Q ˙ x + d Qx dx Q dx where, in this case, heat transfer from the edges is included, P = 2(L + ) is the perimeter of the fin cross section, and Pd x is the surface area of fin available for convection. Introducing Fourier’s law and noting that the area, A, normal to the conduction heat flow path is A = L, we have ˙ x = −k AdT = −kL dT Q dx dx

Steady-State Conduction

643

(a)

(b)

(c) FIGURE 20.29 Extended surfaces: (a) the longitudinal fin of uniform thickness or rectangular profile, (b) the cylindrical spine, and (c) the annular or radial when fin of uniform thickness.

h

T∞ T

Tb

δ dx

k L b x

FIGURE 20.30 Coordinate system for the longitudinal fin of uniform thickness.

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Introduction to Thermal and Fluid Engineering

Then, the energy balance gives rise to the governing differential equation d2T hP − (T − T∞ ) = 0 2 dx kL

(20.53)

We let the temperature excess be = T − T∞ and with

 m=

hP kL

1/2

 ≈

(20.54a)

2h k

1/2 (20.54b)

we may write Equation 20.53 as d 2 dx

2

− m2 = 0

(20.55)

which has a general solution that may be written in terms of exponential functions = C1 e mx + C2 e −mx

(20.56)

= C1 cosh mx + C2 sinh mx

(20.57)

or hyperbolic functions

The evaluation of the constants, C1 and C2 , requires the specification of two boundary conditions, one at the fin base and one at the fin tip. Depending on the operational conditions, a variety of boundary conditions may be appropriate. For example, the temperature at the base may be assumed constant and equal to that of the primary or base surface to which the fin is attached. This implies zero contact resistance between the primary surface and the base of the fin. It is also implicit that the attachment of the fin does not cause a base temperature depression relative to the unfinned portion of the primary surface. Likewise, we may envision several types of boundary conditions at the fin tip. The two considered here are a convective heat loss from the tip, which removes one of the MurrayGardner conditions from consideration, and an insulated tip. 20.6.2.1 Constant Base Temperature with Tip Heat Loss The boundary conditions are (x = 0) = b = Tb − T∞

and

 d  − kL  = ha dx x=b

(20.58)

where Tb is the base temperature and h a is the heat transfer coefficient at the tip. The solution for is = b

cosh m(b − x) + H sinh m(b − x) cosh mb + H sinh mb

(20.59)

where H = h a /km. Because all of the heat dissipated by the fin must, in the steady state, pass the base of the fin, it follows that     d sinh mb + H cosh mb ˙ = −k A  Q = kmL (20.60) b dx x=0 cosh mb + H sinh mb

Steady-State Conduction

645

20.6.2.2 Constant Base Temperature with Insulated Tip Here, the boundary conditions are (x = 0) = b = Tb − T∞

and

 d  kL  =0 dx x=b

(20.61)

˙ are immediately obtainable by setting H = 0. The solutions for and Q = b and

cosh m(b − x) cosh mb

  ˙ = −k Ad  Q = kmL b tanh mb dx x=0

(20.62)

(20.63)

20.6.3 Fin Performance Criteria Three criteria are commonly employed to assess the performance of a fin. The first one is the fin effectiveness, , which is defined as the ratio of the rate of heat dissipated by the fin to the rate of heat transfer from the prime surface if the fin were not present =

˙ Q h A b

(20.64)

If the use of the fin is to be economically justifiable,  should be greater than unity and as large as possible. For the case of the longitudinal fin of constant thickness with a constant base temperature and an insulated tip,   k P 1/2 = tanh mb (20.65) Ah The second performance criterion involves the use of the fin efficiency, , which is ˙ the heat dissipated by the fin to Q ˙ id , and the rate of heat transfer defined as the ratio of Q, from an ideal fin when operating at the base temperature, b , (a fin with infinite thermal conductivity). The fin surface area is S = Pb so that we may express the ideal heat flow from the fin as ˙ id = h Pb b Q

(20.66)

Then, the efficiency of the longitudinal fin with uniform thickness and an insulated tip where the fin surface is S f = 2b L will be the ratio of Equation 20.63 to Equation 20.66

=

˙ Q kL b m tanh mb tanh mb = = ˙ id 2b L b mb Q

(20.67)

The third criterion of performance may be related to the thermal resistance of the fin, which can be expressed as Rf =

b ˙ Q

(20.68)

For a fin to be effective, its thermal resistance must be less than the thermal resistance of the exposed base, 1/ h A = 1/ hL. Thus, Rf =

b 1 < ˙ hL Q

(20.69)

646

Introduction to Thermal and Fluid Engineering T∞ = 20°C h = 50 W/m2-K

Assumed Insulation

250 mm 8 mm k = 35 W/m-K Tb = 100°C

100 mm

FIGURE 20.31 Longitudinal fin of uniform thickness for Example 20.9.

Example 20.9 A longitudinal fin of uniform thickness (Figure 20.31) operates in a con-

vective environment at a temperature of 20◦ C, which provides a heat transfer coefficient of 50 W/m2 -K. The base temperature of the fin is 100◦ C, and the fin is fabricated from steel with k = 35 W/m-K. The fin is 8-mm thick, 100-mm high, and 250-mm long and the fin tip may be assumed to be insulated. Determine (a) the fin tip temperature, (b) the fin heat dissipation, (c) the fin effectiveness, (d) the fin efficiency, and (e) the thermal resistance of the fin.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (5) The Murray-Gardner assumptions apply. (6) The fin tip is insulated. We first calculate, b , A, m, and mb. b = Tb − T∞ = 100◦ C − 20◦ C = 80◦ C A = L = (0.250 m)(0.008 m) = 2.00 × 10−3 m2  m=

2h k



1/2 =

2(50 W/m2 -K) (35 W/m-K)(0.008 m)

1/2 = 18.898 m−1

and mb = (18.898 m−1 )(0.100 m) = 1.89 (a) We use Equation 20.62 to obtain the tip temperature at x = b = 0.100 m (x) = b (x = b = 0.10 m) = =

cosh m(b − x) cosh mb

80◦ C cosh(1.89) 80◦ C = 23.6◦ C 3.3852

Steady-State Conduction

647

and T(x = b) = 20◦ C + 23.6◦ C = 43.6◦ C ⇐ (b) The fin heat dissipation is found from Equation 20.64 ˙ = k A b m tanh mb Q = (35 W/m-K)(0.002 m2 )(80◦ C)(18.898 m−1 ) tanh(1.89) = 101.1 W ⇐ (c) The effectiveness of the fin is given by Equation 20.64 =

˙ Q 101.1 W = = 12.64 ⇐ 2 h A b (50 W/m -K)(0.002 m2 )(80◦ C)

(d) The efficiency may be obtained from Equation 20.67

=

tanh mb tanh(1.89) = = 0.506 ⇐ mb 1.89

(e) The thermal resistance of the fin can be obtained from Equation 20.68 Rf =

b 80◦ C = = 0.792 K/W ⇐ ˙b 101.1 W Q

20.6.4 The Cylindrical Spine or Pin Fin Because the longitudinal fin of uniform thickness (Figure 20.29a) and the cylindrical spine (Figure 20.29b) are both fins of a constant cross section, the equations developed for the longitudinal fin of uniform thickness may be employed for the analysis of the cylindrical spine. The only difference is in the surface and cross-sectional area that may be taken as  S = d and A = d2 (20.70) 4 and

 m=

4h kd

1/2

 =

hP kA

1/2

Example 20.10 A 20-mm-square chip surface is extended by installing 16 aluminum (k = 238 W/m-K) pin fins forming an aligned array as in Figure 20.32. Each pin fin has a diameter of 2 mm and a height of 14 mm. The assembly is cooled by a fan blowing air at 20◦ C and providing a heat transfer coefficient of 100 W/m2 -K. Assuming that the pin fin tips are adiabatic and the chip surface temperature will not exceed 70◦ C, determine the maximum dissipation capability of the chip.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional.

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Introduction to Thermal and Fluid Engineering

20 cm 20 cm

Spine Detail 2 mm 70°C

14 mm k = 238 W/m-K h = 100 W/m2-K

°C

r Ai

20

(a)

(b)

FIGURE 20.32 Array of cylindrical spines for Example 20.10 (a) arrangement and (b) detail.

(3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (5) There is no contact resistance between the finned array and the chip surface. (6) The Murray-Gardner assumptions apply. We first calculate, b , A, m, and mb for each fin. b = Tb − T∞ = 70◦ C − 20◦ C = 50◦ C  A = d 2 = (0.7854)(0.002 m) 2 = 3.14 × 10−6 m2 4

1/2  1/2 4h 4(100 W/m2 -K) m= = = 28.99 m−1 kd (238 W/m-K)(0.002 m) and mb = (28.99 m−1 )(0.014 m) = 0.406 For n pin fins, Equation 20.63 may be adjusted to give with 16 fins ˙ b = nk A b m tanh mb Q = (16)(238 W/m-K)(3.14 × 10−6 m2 )(50◦ C)(28.99 m−1 ) tanh(0.406) = 6.68 W ⇐ The unfinned or prime surface area may be calculated by subtracting the total footprint area of the 16 pin fins from the total chip surface area. Calling this area, Ap , we have Ap = (0.02 m) 2 − 16(3.14 × 10−6 m2 ) = 3.50 × 10−4 m2

Steady-State Conduction

649

and the heat dissipation from the prime or unfinned surface will be ˙ p = h Ap b = (100 W/m2 -K)(3.50 × 10−4 m2 )(50◦ C) = 1.75 W Q ˙ b and Q ˙ p will provide the heat dissipation capability of the chip The sum of Q ˙ =Q ˙b+Q ˙ p = 6.68 W + 1.75 W = 8.43 W ⇐ Q 20.6.5 Annular or Radial Fin of Uniform Thickness The differential equation that governs the temperature profile in the annular or radial fin of uniform thickness (Figure 20.30c) is Bessel’s modified differential equation. The solution to this equation for the temperature profile as well as the fin heat dissipation and its efficiency are in terms of the modified Bessel functions and it is felt that detailing these solutions here will serve no useful purpose. However, these solutions are available in some of the heat transfer textbooks and in Kraus et al. (2001). Figure 20.33 is a graph of the fin efficiency, , plotted as a function of b(2h/k) 1/2 with b = ra − rb for selected values of the radius ratio, ra /rb . Once is read from the graph, the heat dissipation from the fin can be determined from ˙ b = Q ˙ id = h[2(ra2 − rb2 ) + 2ra ](Tb − T∞ ) Q

(20.71)

where 2(ra2 − rb2 ) represents the surface area of the two faces of the fin and 2ra  is the tip surface of the fin.

Example 20.11 The cylinder of an engine made of cast aluminum (k = 174 W/m-K) has 30 equally spaced radial fins (two of which are shown in Figure 20.34). The cylinder has a diameter of 14 cm and is 20-cm long. Each fin is 2-mm thick and 30-mm high. The outer surface of the cylinder attains a temperature of 120◦ C and the air flow over the fins produces

1.0 0.9

Fin Efficiency, η

0.8

ra/rb

δ

0.7 rb

1 1.5 2 3 4 5

0.6 0.5 0.4

b ra

0.3 0.2 0.1 0

0

1.0

2.0

3.0

2h Fin Parameter, m = b kδ

4.0 1/2

FIGURE 20.33 Efficiency of an annular fin of uniform thickness. (From Gardner, 1945.)

5.0

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Introduction to Thermal and Fluid Engineering

30 Fins

FIGURE 20.34 Array of radial or annular fins for Example 20.11. Observe that the entire array contains 30 fins.

a heat transfer coefficient of 40 W/m2 -K. Determine the heat transfer from the cylinder to the air.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction and convection are the modes of heat transfer. (5) The Murray-Gardner assumptions apply. The outer radius of the fin is ra = rb + 3.00 cm = 7.00 cm + 3.00 cm = 10 cm Hence, with b = ra − rb = 3.00 cm  b

2h k

1/2



2(40 W/m2 -K) = (0.030 m) (174 W/m-K)(0.002 m)

1/2 = 0.455

and ra 10 cm = = 1.429 rb 7 cm Figure 20.33 may then be employed to show that the fin efficiency, , will be

= 0.93

Steady-State Conduction

651

For n = 30 fins, Equation 20.71 can be employed to obtain the heat dissipation of the fins with b = 120◦ C − 20◦ C = 100◦ C and ra2 − rb2 = 0.0051 m2 ˙ b = n h[2(ra2 − rb2 ) + 2ra ](Tb − T∞ ) Q = (30)(0.93)(40 W/m2 -K) × 2{0.0051 m2 + (0.10 m)(0.002 m)](120◦ C − 20◦ C) = 3716 W

(3.716 kW)

Next, we determine the heat transferred from the unfinned portion of the cylinder. The prime surface area is Ap = 2rb (L − n) = 2(0.07 m)[0.20 m − 30(0.002 m)] = 0.0616 m2 and ˙ p = h Ap (Tb − T∞ ) = (40 W/m2 -K)(0.0616 m2 )(100◦ C) = 246 W Q The total heat transfer is ˙ =Q ˙b+Q ˙ p = 3716 W + 246 W = 3962 W Q

(3.962 kW) ⇐

We note that the fins provide about 96% of the total heat dissipation.

20.7

Two-Dimensional Conduction

20.7.1 Introduction In steady two-dimensional conduction, the temperature is a function of two spatial coordinates. In Cartesian coordinates where there is no heat generation, a form of the Laplace equation given by Equation 20.10 pertains ∂2 T ∂x2

+

∂2 T ∂ y2

=0

(20.72)

The solution to this equation requires four boundary conditions and for the rectangular plate of thickness  (Figure 20.35) subjected to a sinusoidal temperature variation on the top edge, these boundary conditions are T(0, y) = T1

(left edge)

T(L , y) = T1

(right edge)

T(x, 0) = T1

(bottom edge)

T(x, H) = T1 + a sin(x/L)

(top edge)

652

Introduction to Thermal and Fluid Engineering T1 + a sin πx L

y

T1

H T1

T1

x L FIGURE 20.35 Rectangular plate subject to a sinusoidal temperature variation at its top edge.

20.7.2 Solution Methods Equation 20.72 can be solved using its associated boundary conditions either by an exact analytical method such as the method of separation of the variables or by an approximate analytical method such as the integral method. Numerical methods such as the finite difference method or the finite element method are also available. A discussion of these methods may be found in many of the heat transfer textbooks such as Bejan (1993), Cengel (2003), Holman (2002), Incropera and DeWitt (2003), and Mills (1999). The method of separation of the variables applied to the rectangular plate of Figure 20.35 subjected to the specified boundary conditions gives the solution T(x, y) = T1 +

a sinh(y/L) sin(x/L) sinh(H/L)

20.7.3 The Conduction Shape Factor Method The conduction shape factor method provides a simple equation for the two-dimensional rate of heat transfer ˙ = k S(T1 − T2 ) Q

(20.73)

where S (which equals A/L for a plane wall) is the conduction shape factor and is a function of the two-dimensional geometry. A thermal resistance for a two-dimensional geometry may be identified by writing Equation 20.73 as Rcond =

T1 − T2 1 = ˙ kS Q

(20.74)

We take note of the fact that the conduction shape factor method does not give the temperature distribution in the configuration.

Steady-State Conduction

653

The conduction shape factor has been worked out for a number of configurations. These are either derived from the exact analytical solutions or approximate analytical or numerical solutions. Table 20.2 provides a brief compilation of conduction shape factors for some useful two-dimensional geometries.

TABLE 20.2

Some Conduction Shape Factors (1) Isothermal cylinder of length, L, buried in a semi-infinite solid:

T2

T1 2L S= cosh−1 (2d/z)  2L S= ln 4z/d

L >> z

z

L >> x

d

z > 3d/2

L

(2) Square of side b and length, L centered in square solid of side, a , and length, L:

T2

2L 0.785 ln a /b 2L S= 0.970 ln a /b − 0.50

a /b < 1.4

a

b

S=

a /b > 1.4

T1

(3) Cylinder of length, L and diameter, d, centered in square solid of side, a , and length, L:

T2

a

a >d

d



L >> d

T1

S=

2L ln 1.08 a /d

(4) Isothermal sphere buried in a semi-infinite solid

T2 d

z z > d/2

T1

S=

2L 1 − d/4z

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Introduction to Thermal and Fluid Engineering

TABLE 20.2

Some Conduction Shape Factors (Continued.) (5) Two cylinders of diameters, d1 and d2 and length, L, placed horizontally in an infinite medium.

z T1

T2

d1

⎧ ⎪ ⎪ ⎨

L >> d1 L >> d2 L >> z

⎪ ⎪ ⎩

S= cosh

−1

2L 

4z2 −d12 −d22 2L



d2

(6) Horizontal cylinder of diameter, d and length, L, placed midway between two plates of equal length and infinite width.

T2 z

z >> d/2

d

L >> z

 S=

2L ln(8z/d)

z T1

Example 20.12 A long, steel pipe (k1 = 35 W/m-K) having an inside radius of r1 = 10 cm

and a thickness of 6 mm is laid under the ground (k3 = 0.52 W/m-K) such that the centerline of the pipe is 50 cm below the surface of the ground (Figure 20.36). The pipe carries a hot fluid at 120◦ C that provides a heat transfer coefficient of 80 W/m2 -K. The pipe is covered with a 10-mm-thick layer of calcium silicate (k2 = 0.056 W/m-K). The ground surface is at 20◦ C; determine the heat loss per meter of pipe. TS = 20°C

Soil, k3 = 0.52 W/m-K

50 cm k2 = 0.056 W/m-K k1 = 35 W/m2-K

Hot Fluid T = 120°C h= 80 W/m-K

FIGURE 20.36 Buried pipe for Example 20.12.

20 cm

Steady-State Conduction

655 1 h1A1

ln(r2/r1) 2пk1L

ln(r3/r2) 2пk2L

l Sk3

+ 420°C – 20°C

–

FIGURE 20.37 Electrothermal analog network for Example 20.13.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The heat flow is one dimensional. (3) Thermal properties do not vary with temperature. (4) Conduction is the only mode of heat transfer. (5) There is no contact resistance in this system. The system may be represented by an electrothermal analog network consisting of the inside convection resistance of the pipe, the conduction resistance of the pipe, the conduction resistance of the calcium silicate layer, and the two-dimensional conduction resistance between the outside surface of the pipe and the ground surface. In this network (Figure 20.37), the outside radius of the pipe, and the inside radius of the calcium silicate), r2 , is 10.6 cm, and the outside radius of the calcium silicate, r3 = 11.6 cm. We may calculate three of the four thermal resistances. The convection resistance will be R1 =

1 1 1 = = = 0.0199 K/W 2 h A1 h(2r1 L) (80 W/m -K)(2)(0.1 m)(1 m)

The resistance of the pipe wall will be R2 =

ln(r2 /r1 ) ln(0.106 m/0.100 m) = 2k1 L (2)(35 W/m-K)(1 m)

or R2 =

ln 1.06 0.0583 = = 2.650 × 10−4 K/W 220 W/K 220 W/K

The resistance of the calcium silicate layer is R3 =

ln(r3 /r2 ) ln(0.116 m/0.106 m) = 2k2 L (2)(0.056 W/m-K)(1 m)

or R3 =

ln 1.0943 0.0902 = = 0.256 K/W 0.352 W/K 0.352 W/K

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Introduction to Thermal and Fluid Engineering

We determine R4 = 1/Sk3 where we note that configuration 1 (Table 20.2) represents the problem at hand. The condition that z > 3r3 is satisfied and the shape factor, S, is given by S=

2L ln(2z/r3 )

Thus, S=

2(1 m) 2 m 2 m = = = 2.9170 m ln[4(0.50 m)/(0.232 m)] ln(8.6206) 2.154

Then R4 =

1 1 = = 0.6593 K/W Sk3 (2.9170 m)(0.52 W/m-K)

˙ as and we can now calculate the heat loss, Q, ˙ = Q =

120◦ C − 20◦ C (0.0199 + 2.650 × 10−4 + 0.2561 + 0.6593) 100◦ C 0.9355

= 106.9 W

20.8

Summary

The general equation of heat conduction in rectangular coordinates ∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

+

qi 1 ∂T = k  ∂t

(20.6)

can, in the steady state, and, in the absence of internal volumetric heat generation, be reduced to d2T =0 d x2 in rectangular coordinates and in the x-coordinate direction,   d dT r =0 dr dr in cylindrical coordinates in the r -coordinate direction and   d 2 dT r =0 dr dr

(20.12)

(20.20)

(20.27)

in spherical coordinates in the r -coordinate direction. The critical radius of insulation in a cylindrical or spherical system pertains to the insulation and yields the minimum resistance to heat flow. It is given by rcr,cyl =

k h

(20.25)

Steady-State Conduction

657

for the cylinder and rcr,sphere =

2k h

(20.26)

for the sphere. With a prescribed volumetric heat generation, q i , Equation 20.6 reduces (in the x direction) to d2T qi 1 ∂T + = 2 dx k  ∂t

(20.33)

Extended surfaces or fins are used to augment the heat dissipating capability of a surface. For the longitudinal fin of uniform thickness with an insulated tip, the temperature excess, = T − T∞ , at any point is given by = b

cosh m(b − x) cosh mb

(20.62)

where m = (2h/k) 1/2 . The heat dissipated by the fin is ˙ = kmL b tanh mb Q

(20.63)

The performance of the cylindrical spine is identical to that for the longitudinal fin of uniform thickness except that m = (4h/kd) 1/2 . The fin efficiency in both cases is given by

=

tanh mb mb

(20.67)

The efficiency of the annular or radial fin of uniform thickness may be found from Figure 20.33 and the heat dissipation is given by ˙ b = Q ˙ id = h[2(ra2 − rb2 ) + 2ra ](Tb − T∞ ) Q

(20.71)

Conduction shape factors may be employed for two-dimensional heat conduction analysis with the heat flow given by ˙ = k S(T1 − T2 ) Q

(20.73)

where S is the shape factor. Several useful shape factors are given in Table 20.2.

20.9

Problems

The Plane Wall 20.1: A metal rod 2.5 cm × 2.5 cm in cross section and 18 cm long is heated at one end and cooled at the other. The heat input at the left end is 2 W and the difference in temperature between the left and right hand ends is 5◦ C. Determine the thermal conductivity of the rod. 20.2: A horizontal steel plate (k = 36 W/m-K) that is 5-cm thick is covered with 1 cm of fiber insulating board (k = 0.050 W/m-K). The temperature of the steel and insulating board surfaces are 150◦ C and 25◦ C, respectively. The surface area of the steel is 0.625 m2 . Determine (a) the heat transmitted through the composite and (b) the interface temperature.

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Introduction to Thermal and Fluid Engineering

20.3: A glass window (k = 1.07 W/m-K) measuring 2 m × 3 m is 0.3175-cm thick. The inside air is at 22.2◦ C and the heat transfer coefficient is 12 W/m2 -K. The outside air is at −23.3◦ C and the heat transfer coefficient is 20 W/m2 -K. Determine (a) the heat loss and (b) the temperature difference across the glass. 20.4: Glass storm windows are used to decrease heat transfer between the inside of a house and the environment. For a storm window with two panes of glass separated by a 7.5-mm air space, use the conditions of Problem 20.3 to determine (a) the heat loss and (b) the temperature drop across the air space. 20.5: A boiler is to be insulated so that its heat loss does not exceed 125 W/m2 of wall surface area. The inside and outside surfaces are to be maintained at 825◦ C and 200◦ C. Determine the thickness of glass wool insulation to be used. 20.6: In Problem 20.5, suppose that the glass wool insulation is exposed to the air at 25◦ C through a heat transfer coefficient of 10 W/m2 -K and the heat flux of 125 W/m2 is to be maintained. Determine (a) the thickness of the insulation and (b) its surface temperature. 20.7: One side of an aluminum alloy (k = 182 W/m-K) block is maintained at 180◦ C. The other side is covered with fiberglass (k = 0.038 W/m-K) that is 1.25-cm thick and whose outside face is maintained at 60◦ C. The total heat flow through this composite is 280 W. Determine the cross-sectional area for heat flow. 20.8: The walls of a house are constructed with a 10-cm layer of brick (k = 0.646 W/m-K), 1.25 cm of celotex (k = 0.046 W/m-K), an air space 9.25-cm thick, another layer of 1.25 cm celotex, and 0.635 cm of wood paneling (k = 0.204 W/m-K). The temperature of the outside surface of the brick is 0◦ C and the inside surface of the wood paneling is 25◦ C. Assume that the air space heat transfer is solely by conduction and determine (a) the heat transferred and (b) the heat transferred if the air space is filled with glass wool (k = 0.040 W/m-K). 20.9: A composite slab contains two layers. Layer 1 is brass, which is 2-cm thick and layer 2 is iron, which is 3-cm thick. The iron is exposed to air at 10◦ C with a heat transfer coefficient of 12 W/m2 -K. Determine the temperature at the brass surface if the brass is exposed to a radiation heat flux of 2000 W/m2 . 20.10: The enclosure of a refrigerator is constructed by sandwiching a 4.8-cm layer of fiberglass insulation (k = 0.038 W/m-K) between two 2.5-mm mild steel plates (k = 42.9 W/m-K). The outside and inside temperatures are respectively 27◦ C and 4◦ C and each convection coefficient on the outside and inside have a value of 5.50 W/m2 -K. Determine the heat gain through 1 m2 of the wall. 20.11: A single-room vacation hut has a floor area of 4 m × 6 m and a roof height of 2.25 m. It has a single storm window in the 6-m wall that measures 1.5 × 1.5 m. The effective L/k ratios for the three surfaces are Item Window Walls Roof

L/k, m2 -K/W 0.1194 0.1923 0.3226

The inside surface of the hut is to be held at 22◦ C when the outside surface exposed to the environment is at 0◦ C, and electricity costs $0.89/kW-h. Determine the cost of heating the hut for a continuous heater operation of 2 weeks.

Steady-State Conduction

659

20.12: Wind blows on the outside of the vacation hut in Problem 20.11 such that the heat transfer coefficient between the outside surface of the hut and the environment at 0◦ C is 16 W/m2 -K. Determine the cost of heating for the 2-week period. 20.13: An aluminum pot (k = 232 W/m-K) having a diameter of 35 cm holds boiling water (100◦ C) and is sitting on a heating element that injects heat into the pot at a rate of 750 W. The wall of the pot is 1.125-mm thick and the temperature at the inside surface of the pot is 102◦ C. Determine (a) the boiling heat transfer coefficient and (b) the temperature at the exterior of the pot just above the heating element. 20.14: The composite wall of a furnace consists of three layers of different materials. Layer 1 (k = 24 W/m-K) is the inner layer and is 30-cm thick. Layer 2 (the middle layer) is 75-cm thick and its thermal conductivity is unknown. Layer 3 (k = 48 W/m-K) is the outer layer and is 15-cm thick. The inside and outside temperatures are 620◦ C and 20◦ C, respectively, and the heat flux is measured as 5 kW/m2 . Determine the thermal conductivity of the middle layer. 20.15: The composite wall of a furnace is configured as follows: Item 1 2 3

Material Fireclay brick Diatomaceous earth brick Common brick

k W/m-K

Thickness, cm

1.07 0.31 0.69

20 15 10

The surface adjacent to the fireclay brick is held at 1075◦ C and the surface adjacent to the common brick is held at 125◦ C. Determine (a) the heat loss per m2 of wall area, (b) the temperatures between the layers, and (c) the temperature at a point 15 cm from the outer surface. 20.16: A slab has a thermal conductivity of 12 W/m-K and a thickness of 10 cm. For unit cross-sectional area and the left-hand and right-hand faces maintained at 130◦ C and 10◦ C, respectively, determine the heat flow from left to right. Contact Resistance 20.17: Reconsider Problem 20.2 with a contact resistance of Rc = 0.0125 m2 -K/W between the steel and the insulating board. Determine (a) the heat transmitted through the composite and (b) the temperature drop due to the contact resistance. 20.18: Suppose that in Problem 20.9, there is a thermal contact resistance of Rc = 0.00875 m2 K/W but all other conditions remain the same. Determine the temperature of the brass surface. 20.19: Reconsider Problem 20.14 and suppose that, along with the resistances given, there is a contact resistance of Rc = 0.0075 m2 -K/W between layers 1 and 2 and a contact resistance of Rc = 0.0065 m2 -K/W between layers 2 and 3 causing the heat loss to be reduced to 4.479 kW. Determine the thermal conductivity of layer 2. 20.20: Two insulated stainless steel cylindrical rods, 10-cm long with k = 12.8 W/m-K are placed end to end. The interface has a contact resistance of Rc = 0.0048 m2 -K/W. The ends of the rod are maintained at 240◦ C and 40◦ C. Determine the diameter of the rods if a heat flow of 50 W is measured. 20.21: An electronic component dissipating 22.5 W is to be conductively cooled via a heat sink having an effective thermal resistance of 0.2431◦ C/W and two contacts, each with Rc = 10−4 m2 -K/W. The area of the contacts is 4.00 cm2 , and the maximum

660

Introduction to Thermal and Fluid Engineering permitted component case temperature is 85◦ C. Determine if the component must be derated if it sits on a surface at 25◦ C.

20.22: Two square copper (k = 385 W/m-K) bars having a length of 22.5 cm and an area of 16 cm2 are pressed together at their ends so that the contact resistance is Rc = 2 × 10−4 m2 -K/W. The overall temperature difference between the ends is 50◦ C. Determine the axial heat flow if (a) the contact resistance is neglected and (b) the contact resistance is included. 20.23: Rework Problem 20.22 with one of the bars specified as aluminum (k = 178 W/m-K) and the other as copper (k = 360 W/m-K), both having a surface roughness of 50 m. The contact pressure is 25 MPa. 20.24: Two 3.5-cm-diameter stainless steel bars (k = 16 W/m-K) having a length of 12.5 cm have a contact between them (Rc = 5.25 × 10−4 m2 -K/W). The overall temperature difference between the ends of the bars is 80◦ C. Determine (a) the axial heat flow and (b) the temperature drop across the contact. Cylindrical Pipes and the Critical Radius of Insulation 20.25: A nominal 5-in schedule 40 cast-iron (k = 46.4 W/m-K) pipe is covered with 4 cm of 85% magnesia (k = 0.074 W/m-K) insulation. The pipe carries steam such that the inner pipe surface is 435◦ C and the outer insulation surface is at 40◦ C. Determine the heat loss for a pipe length of 40 m. 20.26: A nominal 8-in schedule 80 mild steel (k = 39.6 W/m-K) pipe is covered with a 6 cm of 85% magnesia (k = 0.074 W/m-K) insulation and then by a 3-cm layer of an insulation with k = 0.062 W/m-K. The inside pipe surface is at 450◦ C and the outer insulation surface is at 50◦ C. For a pipe length of 25 m, determine (a) the heat lost and (b) the temperature between the insulation layers. 20.27: Suppose that the insulation layers in Problem 20.26 are reversed. Determine the heat lost. 20.28: A pipe in a refrigeration system has an outside diameter of 6 cm and a surface temperature of 4◦ C. It is covered with two layers of insulation, a 2.5-cm layer of 85% magnesia (k = 0.074 W/m-K) and a 2.5-cm layer of rock wool. Determine which layer should be placed adjacent to the pipe if the surface of the outer layer, whichever one, is to be maintained at 36◦ C. 20.29: A steel pipe (k = 25.5 W/m-K) has an outside diameter of 5.50 cm. It is covered with a 0.75-cm layer of insulation (k = 0.172 W/m-K) and 2.5-cm layer of another insulation (k = 0.045 W/m-K). The surface temperature of the pipe is 325◦ C and the outer insulation temperature is 25◦ C. For a length of 2 m, determine (a) the heat loss and (b) the temperature at the interface between the two insulation layers. 20.30: In Problem 20.29, determine the thickness of the outer layer of insulation to limit the heat loss to 240 W. 20.31: A nominal 1-in schedule 40 wrought iron (k = 48 W/m-K) is covered with 1.25 cm of insulation (k = 0.060 W/m-K). The inside surface of the pipe is at 275◦ C and the temperature at the outer surface of the insulation is 50◦ C. The pipe length is 1 m. Determine (a) the heat loss and (b) the temperature at the insulation interface. 20.32: In Problem 20.31, determine the required insulation thickness of the outer layer of insulation to limit the heat loss to 120 W. 20.33: Rework Problem 20.25 with convection on both the inside pipe surface (h = 120 W/m2 -K) and the outer insulation surface (h = 12.5 W/m2 -K). In this case, the temperature of the steam is 450◦ C and the environmental temperature is 20◦ C.

Steady-State Conduction

661

20.34: Rework Problem 20.26 with convection on both the inside pipe surface (h = 600 W/m2 -K) and the outer insulation surface (h = 12 W/m2 -K). In this case, the temperature of the steam is 450◦ C and the environmental temperature is 50◦ C. 20.35: Rework Problem 20.29 with convection on the outer insulation surface (h = 10 W/m2 -K) with an environmental temperature of 25◦ C. All other parameters remain the same. 20.36: Steam at 1.4 MPa flows through an 8-in schedule 80 mild steel pipe that is covered with 2.25 cm of insulation (k = 0.058 W/m-K). The heat transfer coefficient at the inner pipe surface is 400 W/m2 -K and the outside surface is exposed to air at 20◦ C where it loses heat by convection through a heat transfer coefficient of 20 W/m2 -K. The outer surface has an emissivity of 0.78 and the walls of the enclosure are at 20◦ C. For a pipe length of 1.2 m, determine (a) the heat loss and (b) the temperature at the surface of the insulation. 20.37: A 5-cm-outside-diameter tube with a surface temperature of 240◦ C passes through an enclosure where the air temperature is 25◦ C. Insulation (k = 0.205 W/m-K) is to be applied and the natural convection heat transfer coefficient at the surface temperature of the insulation is expected to be h = 4.25 W/m2 -K. Determine (a) the critical radius of the insulation, (b) the heat loss from 1 m of the pipe before application of the insulation, and (c) the heat loss from 1 m of the pipe after application of the insulation. 20.38: A steel tube (k = 44 W/m-K) has an inside diameter of 3.00 cm and a tube wall thickness of 1.5 mm. A fluid at 245◦ C that produces a heat transfer coefficient of 1600 W/m2 -K flows on the inside of the tube and a second fluid flows across the outside of the tube with a heat transfer coefficient of 200 W/m2 -K at 35◦ C. The tube is 1-m long. Determine the heat loss. 20.39: To reduce the heat loss in Problem 20.38, the tube is to be equipped with asbestos insulation. All conditions listed in Problem 20.38 are to remain the same. Determine the thickness of insulation required to reduce the heat loss to 500 W. 20.40: A 6-cm-outside-diameter steel pipe with a surface temperature of 385◦ C is covered with a 1.25-cm layer of an insulation (k = 0.225 W/m-K) followed by a 2-cm layer of another insulation (k = 0.080 W/m-K). The outside of the 2-cm layer is exposed to an environment via a heat transfer coefficient of 25 W/m2 -K, where the air temperature is 25◦ C. Determine (a) the heat loss and (b) the temperature at the interface between the two layers of insulation. Hollow Spheres 20.41: A hollow stainless steel sphere (0.1% chrome) has an outer diameter of 80 cm and an inner diameter of 64 cm. Its inside surface is at 40◦ C and it is placed in a temperature bath at 20◦ C with a heat transfer coefficient of 125 W/m2 -K. Determine (a) the heat flow from the sphere and (b) the temperature at the surface of the sphere. 20.42: Suppose that a 10-cm layer of insulation (k = 0.165 W/m-K) is added to the outside of the sphere in Problem 20.41 and causes the heat transfer coefficient to drop to 120 W/m2 -K. Determine the heat loss. 20.43: A hollow sphere is fabricated from 22% Si aluminum with inner and outer diameters of 4 cm and 10 cm, respectively. The inside and outside temperatures are 60◦ C and 20◦ C. Determine the heat flow from the sphere.

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Introduction to Thermal and Fluid Engineering

20.44: In an effort to reduce the heat loss, the sphere of Problem 20.43 is to be covered with two layers of insulation, each 2-cm thick. Layer 1 has a thermal conductivity of k1 = 0.0625 W/m-K and layer 2 has a thermal conductivity of k2 = 0.125 W/m-K. The surface temperature of layer 2 is 20◦ C and the inside of the sphere remains at 60◦ C. Determine the heat loss. 20.45: Suppose that in Problem 20.44, the two insulation layers are reversed. Determine the heat loss if all other conditions remain the same. 20.46: Reconsider Problem 20.43 with the environment at 20◦ C and a heat transfer coefficient at the exterior of the sphere at 550 W/m2 -K. Determine the heat loss. 20.47: Suppose that in Problem 20.46, the critical radius of an insulation with k = 0.938 is added to the sphere. All other conditions remain the same. Determine the heat loss. 20.48: Rework Problem 20.47 with insulation layers (a) 3-cm thick (k = 0.15 W/m-K) and (b) 4-cm (k = 0.12 W/m-K) thick. 20.49: A sphere with a diameter of 12.5 cm is electrically heated so that its surface temperature is held at 240◦ C. The surface is exposed to an environment at 25◦ C via a heat transfer coefficient of 48 W/m2 -K. Determine the heat loss from the sphere. 20.50: A sphere with a diameter of 12.5 cm is exposed to an environment with a heat transfer coefficient of 12 W/m2 -K and has a surface emissivity of 0.475. Both the enclosure and the environment are at 20◦ C. Determine the surface temperature if the heat loss is 600 W.

Internal Heat Generation 20.51: A plane wall with a thickness of 16 cm and a thermal conductivity of 9.6 W/m-K has a volumetric heat generation of 160 kW/m.3 The left and right sides of the wall are held at 95◦ C and 75◦ C. Determine (a) the location of the maximum temperature in the wall and (b) the maximum temperature. 20.52: A plane wall has a thickness of 4 cm and a center plane temperature of 180◦ C. The wall has a thermal conductivity, k W/m-K, and both faces are maintained at 60◦ C. Determine the uniform heat generation in the wall. 20.53: Consider a plane slab with thermal conductivity, k, and faces held at T1 and T2 . The slab is L m thick and the volumetric heat generation is of q i (W/m3 ). With the temperature distribution of Equation 20.34, show that Equation 20.37 results. 20.54: A plane wall with a thickness of 10 cm and a thermal conductivity of 30 W/m-K experiences a volumetric heat generation of 5 × 105 W/m3 . The left face of the wall is in contact with a coolant at 25◦ C, which provides a heat transfer coefficient of 100 W/m2 -K. The right face of the wall is cooled by a fluid at 30◦ C with a heat transfer coefficient of 300 W/m2 -K. Determine the temperatures of the faces and the maximum temperature. 20.55: A plane fuel element in a nuclear reactor is 3-cm thick and has a thermal conductivity of 60 W/m-K. The element is exposed on both faces to a coolant at 25◦ C that creates a heat transfer coefficient of 150 W/m2 -K. Determine the maximum possible volumetric heat generation in the element if the maximum temperature in the fuel element is not to exceed 58◦ C. 20.56: A 20-cm-thick plane slab has a centerline temperature of 250◦ C and a surface temperature of 25◦ C. Determine the distance from the centerline for a temperature of 160◦ C.

Steady-State Conduction

663

20.57: A current of 180 A is passed through a 2-m-long and 2.5-mm-diameter wire (k = 25 W/m-K and e = 7.2 × 10−8 Om-m). The wire is submerged in a liquid bath at 105◦ C and experiences a convective dissipation with a constant heat transfer coefficient of 3000 W/m2 -K. Determine the center temperature of the wire. 20.58: A 1-cm-diameter electric wire with a resistance of 0.03  per meter of length carries a current of 80 A under steady-state operation. The conductor dissipates the heat generated to its surroundings at 27◦ C, which provides a heat transfer coefficient of 950 W/m2 -K. The thermal conductivity of the wire is 25 W/m-K. Determine (a) the volumetric heat generation in the wire, (b) the wire surface temperature, and (c) the maximum temperature in the wire. 20.59: A solid cylinder with a diameter, do , of 10 cm and thermal conductivity, k = 18 W/m-K has a surface temperature of 96◦ C. Determine the volumetric heat generation if the temperature at a point 2.5 cm from its center is 170◦ C. 20.60: A solid sphere with a radius of 10 cm has a thermal conductivity of 1.2 W/m-K. The center temperature is 240◦ C and the surface temperature is 40◦ C. Determine the volumetric heat generation. 20.61: A solid sphere has a radius of 12 cm. The volumetric heat generation is 600,000 W/m3 and the thermal conductivity of k = 20 W/m-K. The sphere is cooled via a heat transfer coefficient of 200 W/m2 -K in an environment of 20◦ C. Determine (a) the surface temperature, (b) the center temperature, and (c) the temperature at a point 4 cm from the center. 20.62: A solid sphere 24 cm in diameter has a uniform volumetric heat generation of 800,000 W/m3 and a thermal conductivity of 16 W/m-K. The sphere has a maximum temperature of 172◦ C and sits in an environment of 20◦ C. Determine the heat transfer coefficient. Extended Surfaces (Fins) 20.63: A cylindrical rod of pyrex glass with a diameter of 1.375 cm and a base temperature of 125◦ C extends 10 cm into the air at 25◦ C, where the heat transfer coefficient is 10 W/m2 -K. Determine (a) the temperature at the mid-height, (b) the temperature at the tip, (c) the heat dissipated by the rod, (d) the fin effectiveness, (e) the fin efficiency, and (f) the thermal resistance. 20.64: Reconsider Problem 20.63 with a cast-iron rod (k = 50 W/m-K), all other conditions remaining the same. 20.65: Reconsider Problem 20.63 with an aluminum rod (k = 180 W/m-K), all other conditions remaining the same. 20.66: A longitudinal fin of uniform thickness with a base temperature of 140◦ C operates in a convective environment with the air at 15◦ C at a heat transfer coefficient of 50 W/m2 -K. The fin is 4-mm thick, 8-cm high and 40-cm long. The fin is fabricated of a steel having a thermal conductivity k of 27.5 W/m-K. Determine (a) the tip temperature, (b) the heat dissipated by the fin, (c) the fin efficiency, (d) the fin effectiveness, and (e) the thermal resistance. 20.67: Reconsider Problem 20.66 with aluminum (k = 202 W/m-K), all other conditions remaining the same. 20.68: Reconsider Problem 20.66 with copper (k = 385 W/m-K). All other conditions remaining the same.

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20.69: Three cylindrical spines, one of glass (k = 1.12 W/m-K), one of iron (k = 55 W/m-K) and one of aluminum (k = 205 W/m-K), have a diameter of 0.75 cm and a height of 6 cm. Their tips are insulated and the heat transfer coefficient in each case is 12 W/m2 -K. The environmental temperature is 25◦ C and the spine base temperature is 85◦ C. Determine the tip temperature in each case. 20.70: Rework Problem 20.69 if all three spines dissipate from their tips to the environment via a heat transfer coefficient of 10 W/m2 -K. All other conditions remain the same. 20.71: A 1.75-cm-diameter alloy rod (k = 50 W/m-K) with a base temperature of 227◦ C and a height, b, extends into air at 27◦ C such that the heat transfer coefficient is 50 W/m2 -K. Neglecting the fin tip heat loss, determine the height, b, necessary to achieve a tip temperature of 172◦ C. 20.72: A longitudinal fin of uniform thickness has a thermal conductivity of 62.5 W/m2 -K and a thickness of 0.25 cm. It has a base temperature of 80◦ C and extends into the air at 20◦ C such that the heat transfer coefficient is 75 W/m2 -K. The fin length is 40 cm and the fin tip is insulated. Determine the heat flow for fin heights of (a) 1.5 cm, (b) 3 cm, and (c) 5 cm. 20.73: Rework Problem 20.72 for the case of a tip heat loss through a heat transfer coefficient having a value of 50 W/m2 -K. All other conditions remain the same. 20.74: An annular fin of uniform thickness has an inner radius of 2.5 cm and an outer radius of 6 cm. The fin thickness is 2.25 cm and the fin possesses a thermal conductivity of 200 W/m-K. The base temperature is 165◦ C and the fin dissipates to air at 25◦ C with a heat transfer coefficient of 300 W/m2 -K. Determine (a) the fin efficiency, (b) the heat dissipated by the fin, (c) the fin effectiveness, and (d) the thermal resistance of the fin. 20.75: Rework Problem 20.74 for a fin with a thermal conductivity of 100 W/m-K. All other conditions remain the same. 20.76: One end of an aluminum rod (k = 208 W/m-K) is held at 200◦ C. The other end is held at 100◦ C and the rod height is 25 cm. The rod has a diameter of 2.5 cm and the heat transfer coefficient between the rod and the surroundings is 20 W/m2 -K. Determine the net rate of heat loss from the rod. The Shape Factor Method 20.77: A very long pipe with a diameter of 20 cm is buried horizontally in a soil with a thermal conductivity of k = 1.4 W/m-K at a depth of 1 m below the surface. The surface temperatures of the pipe and soil are 80◦ C and 30◦ C, respectively. Determine the heat loss per unit length. 20.78: Two very long pipes with diameters of d1 = 12 cm and d2 = 16 cm are buried horizontally in a soil with a thermal conductivity of k = 1.5 W/m-K at a depth of 6.20 m below the surface. The centerline spacing of the pipes is 1 m and their surfaces are held at T1 = 25◦ C and T2 = 75◦ C. Estimate the heat flow between them. 20.79: A tall chimney with the outside cross section of a square has a side dimension of 1 m. A circular flue in the center of the chimney has a diameter of 20 cm and carries a gas such that the flue surface temperature is 180◦ C. The chimney material has a thermal conductivity of 4 W/m-K and its surface temperature is held at 20◦ C. Determine the heat loss per meter of chimney height.

Steady-State Conduction

665

20.80: For the chimney in Problem 20.79, consider that the heat transfer coefficient between the outside surface and the environment is 1.25 W/m2 -K. The outside surface is at 20◦ C. Determine (a) the heat loss per meter of length and (b) the temperature at the surface of the flue. 20.81: Rework Problem 20.79 for a concentric square flue that is 12 cm × 12 cm. All other conditions remain the same. 20.82: An 8-cm-diameter sphere whose surface temperature is 180◦ C is buried at a depth of 80 cm in soil having a thermal conductivity of 135 W/m-K. The surface of the soil is held at 30◦ C. Determine the heat loss from the sphere. 20.83: A 1.2-m-long 4-cm-diameter pipe is located at the centerline of a plastic (k = 8 W/m-K) slab that is 40-cm thick. If the pipe surface is held at 100◦ C and the slab surface is held at 20◦ C, determine the heat loss from the pipe.

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21 Unsteady-State Conduction

Chapter Objectives •

To describe the physics of unsteady or transient conduction.

To illustrate the use of the lumped capacitance model for the convective cooling of a body of arbitrary shape. • To develop a criterion for the validity of the lumped capacitance model. •

To describe the semi-infinite model and present solutions for three types of surface thermal boundary conditions. • To study one-dimensional convective cooling/heating of a plane wall, a solid cylinder and a solid sphere with the help of simple approximate solutions and the Heisler charts. •

21.1

Introduction

In unsteady or transient conduction the temperature of a system changes with time due to a thermal disturbance created within the system and/or the environment. For example, when a power transistor is suddenly energized, the heat generated within the transistor causes its temperature to increase with time. When a hot billet is withdrawn from a furnace and immersed in a coolant pool, it is the change in the environmental condition that causes the temperature of the billet to change with time. Consider the cooling of a hot solid when it is immersed in a pool of coolant. During the early period, the cooling effect is felt in a thin region of the solid that is closest to the surface. The internal core of the solid is virtually unaffected by the cooling that is occurring at the surface. This early thermal regime that is characterized by steep temperature gradients may be modeled by considering the solid to be semi-infinite in extent. As time progresses, the cooling effect penetrates deeper and deeper into the solid and the temperature gradients begin to moderate. As the late stages of the transient are reached, temperature gradients become so small that the body temperature changes with time but is essentially uniform throughout the solid. It is this behavior that is characterized by the lumped capacitance model. This model is the easiest to analyze and is considered first.

667

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Introduction to Thermal and Fluid Engineering

The Lumped Capacitance Model

The lumped capacitance model ignores the spatial temperature variation in the body and assumes that the temperature is a function only of time. Despite its simplicity, the model serves as a useful predictive tool for studying many practical problems involving transient conduction. 21.2.1 Convective Cooling Consider a solid body of arbitrary shape having volume, V, surface area, S, density, , and specific heat, c. The body is initially at a uniform temperature, Ti , as shown in Figure 21.1. At time, t ≥ 0, the body is immersed in a coolant, which is at T∞ < Ti . The coolant is assumed to have an infinite thermal capacity so that its temperature is unaffected by any energy that is released from the solid body. The energy balance dictates that the rate of surface convection must equal the rate of decrease of the internal energy of the solid body, that is −Vc

dT = hS(T − T∞ ) dt

(21.1)

where the minus sign has been inserted because the temperature gradient, dT/dt, is less than zero. The initial condition may be expressed as T(t = 0) = Ti

(21.2)

To solve Equation 21.1, we separate the variables to yield dT hS =− dt T − T∞ Vc

(21.3)

and then an integration gives ln(T − T∞ ) = −

hS t + C1 Vc

(21.4)

Surface Area, S

Density, ρ Specific Heat, c Ti Volume, V

Coolant at T∞< Ti

Heat Transfer Coefficient, h

FIGURE 21.1 Solid body of arbitrary shape used to develop the lumped capacitance model for unsteady-state heat conduction.

Unsteady-State Conduction

669

where C1 is the constant of integration that is found by applying the initial condition of Equation 21.2 C1 = ln(Ti − T∞ ) With the substitution of C1 , Equation 21.4 may be written as T − T∞ = e −(hS/Vc)t Ti − T∞

(21.5)

The quantity, Vc/hS, is the product of the convection resistance, 1/hS, and, Vc, a parameter known as the thermal capacitance of the body. If we draw the analogy with the time constant of a simple first-order electrical network containing a resistor and a capacitor, we see that the quantity, Vc/hS, may be called the thermal time constant, which we denote as .  = Vc/hS

(21.6)

We note that the larger the value of  resulting from a larger thermal capacitance and/or the smaller the product hS, the slower the cooling of the body and the longer the body will take to reach thermal equilibrium with the coolant. The cumulative heat loss to the coolant, Q, may be obtained by integrating the instantaneous convective heat loss  t Q= hS(T − T∞ )dt (21.7) 0

and with the substitution of T − T∞ from Equation 21.5, we may perform the integration to obtain     Q = Vc(Ti − T∞ ) 1 − e −(hS/Vc)t = Vc(Ti − T∞ ) 1 − e −t/ (21.8) 21.2.2 The Validity Criterion To establish the criterion for the validity of the lumped capacitance model, consider onedimensional steady conduction in the plane wall of area, A, thickness, L, and thermal conductivity, k. Let the left face of the wall be held at a fixed temperature while the right face loses heat by convection to a fluid with a convection heat transfer coefficient, h. Because the temperature drops in the wall, Twall , and the fluid, Tfluid , are respectively proportional to the conduction and convection resistances, L/k A, and 1/ h A, where A = S, it follows that Twall L/k A h L = = Tfluid 1/ h A k

(21.9)

The ratio of the conduction resistance to the convection resistance is called the Biot number or Biot modulus and denoted by Bi. Equation 21.9 may be written as Twall = (Bi)Tfluid

(21.10)

and if Bi is much less than 1, Equation 21.10 indicates that Twall will be much less than Tfluid and, therefore, the temperature in the wall may be assumed to be nearly uniform. Now consider a transient conduction situation where a body, initially at a uniform temperature, is suddenly exposed to a convective environment. If the condition of Bi  1 is satisfied, then the transient conduction in the body can be modeled as a sequence of thermal

670

Introduction to Thermal and Fluid Engineering 20 W/m2-K

c = 480J/kg-K ρ = 8055 kg/m3 K = 15.1 W/m-K Ti = 1000 K

10 mm

T∞ = 300 K

FIGURE 21.2 Configuration for Example 21.1.

states where the temperature of the body changes with time but remains spatially uniform at every instant of time. These are precisely the conditions for the lumped capacitance model and the criterion for its validity may be set as Bi  1. The threshold value of Bi that is commonly used is 0.1 so that the lumped capacitance model is valid if Bi =

h Lc < 0.1 k

(21.11)

where L c = V/S is the characteristic length of the body. For a long cylinder, Lc =

V ro2 L ro = = S 2ro L 2

Lc =

V 4ro3 /3 ro = = 2 S 4ro 3

and for a sphere

Example 21.1 A 10-mm-diameter AISI 302 stainless steel ball (c = 480 J/kg-K,  = 8055 kg/m3 , and k = 15.1 W/m-K) is annealed by heating it to 1000 K in a furnace and then allowed to cool to 400 K in the air at 300 K, which provides a convective heat transfer coefficient of 20 W/m2 -K. Determine (a) the time required to cool the ball and (b) the cumulative energy loss over the time determined in (a).

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (a) We first determine the Biot number Bi =

hro (20 W/m2 -K)(0.005 m) = = 2.21 × 10−3 3k 3(15.1 W/m-K)

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671

Because Bi < 0.1, the lumped capacitance model may be employed. Equation 21.5 may be solved to yield     Vc T − T∞ T − T∞ t=− ln = − ln hS Ti − T∞ Ti − T∞ With the time constant, , Vc (8055 kg/m3 )(4/3)(0.005 m) 3 (480J/kg-K) = = 322.2 s hS (20 W/m2 -K)(4)(0.005 m) 2 we have

 t = −(322.2 s) ln

400 K − 300 K 1000 K − 300 K

 = −(322.2 s) ln (0.143)

= −(322.2 s)(−1.946) = 627 s

(10.45 min) ⇐

(b) Using Equation 21.8, the cumulative heat loss is   Q = Vc(Ti − T∞ ) 1 − e −t/ With Vc = (8055 kg/m3 )(4/3)(0.005 m) 3 (480 J/kg-K) = 2.024 J/K and t 627s = = 1.946  322.2s we have Q = (2.024 J/K)(1000 K − 300 K)(1 − e −1.946 ) = (1417.1 J)(1 − e −1.946 ) = (1417.1 J)(0.857) = 1214.7 J ⇐ 21.2.3 The Effect of Internal Heat Generation When an internal heat generation is present, as in the case of an electronic component that is suddenly energized, a heat generation term, E˙ g (in watts), is included in Equation 21.1. With E˙ g included and noting that dT/dt > 0 for this case of heating, Equation 21.1 is modified to Vc

dT = E˙ g − hS(T − T∞ ) dt

(21.12)

The solution to Equation 21.12 is facilitated by introducing a new variable, , defined as  = E˙ g − hS(T − T∞ ) so that d dT = −hS dt dt

(21.13)

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or dT 1 d =− dt hS dt Then, Equation 21.12 may be written as −

Vc d = hS dt

and after the variables are separated d hS =− dt  Vc Integration then gives ln  = −

hS t t + C1 = − + C1 Vc 

(21.14)

The initial condition of Equation 21.2 translates to the condition (t = 0) = E˙ g − hS(Ti − T∞ )

(21.15)

Hence, C1 = ln[ E˙ g − hS(Ti − T∞ )] and with the substitution of this into Equation 21.14 we obtain E˙ g − hS(T − T∞ ) = e −t/ E˙ g − hS(Ti − T∞ )

(21.16)

The steady-state temperature, Tss , may be found by letting t −→ ∞. Thus, with t = ∞ in Equation 21.16 Tss = T∞ +

E˙ g hS

(21.17)

Example 21.2 An electronic component, which has a heat capacity of Vc = 60 J/K, generates 8 W when it is suddenly energized. The component, sketched in Figure 21.3, is initially at a temperature of 25◦ C and a stream of cooling air at 20◦ C provides a heat transfer coefficient of 100 W/m2 -K as soon as the power is applied. The surface area exposed to the cooling air is 50 cm2 . Assume that the lumped capacitance model is valid and determine ρVc = 50 J/K Ėg = 8 W Ti = 25°C Cooling Air T∞ = 20°C h = 100 W/m2-K

FIGURE 21.3 Configuration for Example 21.2.

S = 50 cm2

Unsteady-State Conduction

673

(a) the component temperature after one minute of operation and (b) the steady-state temperature of the component.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (4) Use of the lumped capacitance model is specified. (a) Equation 21.16 may be used to determine the temperature after 1 min (60 s). Here, the time constant is =

Vc 60 J/K = = 120 s hS (100 W/m2 -K)(0.005 m2 )

so that t 60 s = = 0.50  120 s Then via Equation 21.16 with hS = (100 W/m2 -K)(0.005 m2 ) = 0.5 W/K E˙ g − hS(T − T∞ ) = e −t/ E˙ g − hS(Ti − T∞ ) 8 W − (0.5 W/K)(T − 20◦ C) = e −0.50 = 0.607 8 W − 12.5 W + 10 W 8 W − (0.5 W/K)(T − 20◦ C) = 0.607(5.50 W) −(0.5 W/K)(T − 20◦ C) = −4.662 W T − 20◦ C = 9.32◦ C T = 29.3◦ C ⇐ (b) The steady-state operating temperature is given by Equation 21.17 Tss = T∞ +

21.3

E˙ g 8W = 20◦ C + = 36◦ C ⇐ hS 0.50 W/K

The Semi-Infinite Solid

As illustrated in Figure 21.4, the semi-infinite model treats the solid as extending to infinity in all dimensions except for one surface where a thermal boundary condition may be imposed. The temperature distribution in the solid is a function of the spatial coordinate, x, and time, t, that is T(x, t), and it is governed by the Fourier equation in one dimension

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α k

x

FIGURE 21.4 Semi-infinite solid. All dimensions except x extend to infinity.

given by Equation 20.8 ∂2 T ∂x2

=

1 ∂T  ∂t

(21.18)

It will be assumed that the solid is initially at a uniform temperature, Ti , throughout its extent and that the thermal boundary condition at x = 0 has no impact at large distances from the surface. These assumptions can be represented as T(x, 0) = Ti

(21.19)

T(∞, t) = Ti

(21.20)

and

The three types of boundary conditions may be considered for the surface at x = 0 are discussed in Sections 21.3.1 through 21.3.3. 21.3.1 Constant Surface Temperature The case of constant surface temperature is represented by T(0, t) = To

(21.21)

T(x, t) = Ti + (To − Ti ) erfc()

(21.22)

for which the solution for T will be

where =

x 2(t) 1/2

(21.23)

and where erfc denotes the tabulated mathematical relationship known as complementary error function  x 2 2 erfc(x) = 1 − √ e −u du (21.24)  0

Unsteady-State Conduction

675

In this case, the surface heat flux q˙ o = −k ∂x ∂T x=0

is given by q˙ o =

k(To − Ti ) (t) 1/2

(21.25)

21.3.2 Constant Surface Heat Flux The case of constant surface heat flux is represented by q˙ o = −k

∂T ∂x

(0, t)

(specified)

(21.26)

and the solution for T will be T(x, t) = Ti +

2q˙ o (t/) 1/2 −x2 /4t q˙ o x e − erfc() k k

(21.27)

Table 21.1 gives values of the complementary error function for arguments between 0.00 and 3.00. 21.3.3 Surface Convection Here, the surface condition is −k

∂T ∂x

(0, t) = h[T∞ − T(0, t)]

(21.28)

and the solution for T will be T(x, t) = Ti + (T∞ − Ti )

    h√ (hx/k)+(h 2 t/k 2 ) × erfc() × e erfc t +  k TABLE 21.1

Values of the Complementary Error Function, erfc(z) z

erfc(z)

z

erfc(z)

z

erfc(z) × 103

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

1.0000 0.8875 0.7773 0.6714 0.5716 0.4795 0.3961 0.3222 0.2579 0.2031 0.1573

1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00

0.1198 0.0897 0.0660 0.0477 0.0339 0.0237 0.0162 0.0109 0.0072 0.0047

2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00

2.980 1.863 1.143 0.689 0.407 0.236 0.134 0.075 0.041 0.022

(21.29)

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Example 21.3 A thick steel slab has a thermal conductivity k of 13.4 W/m-K and a ther-

mal diffusivity  of 3.70 × 10−6 m2 /s. The slab has a uniform initial temperature of 27◦ C. Determine the temperature at a depth of 2 cm and time t = 5 min if (a) the surface (x = 0) is suddenly changed to and kept at 50◦ C, and (b) the surface (x = 0) is heated with a burner that provides a surface heat flux of 10,000 W/m2 . Then, determine the cumulative energy supplied over the time period.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (4) A semi-infinite solid is presumed. (a) We first determine the quantity, , given by Equation 21.23 =

x 0.02 m = = 0.30 1/2 −6 2(t) 2[(3.70 × 10 m2 /s)(300 s)]1/2

and from Table 21.1, we find that erfc(0.30) = 0.6714. Then Equation 21.22 may be used to obtain T at x = 0.02 m and t = 5 min or 300 s T(x, t) = Ti + (To − Ti ) erfc() T(0.02 m, 300 s) = 27◦ C + (50◦ C − 27◦ C) erfc(0.30) = 27◦ C + (23◦ C)(0.6714) = 27◦ C + 15.4◦ C = 42.4◦ C ⇐

x=0 T(x10) = 27°C k = 13.4 W/m-K α = 370 × 10—6 m2/s x=2m

x FIGURE 21.5 Configuration for Example 21.3.

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677

The cumulative energy supplied can be determined by integrating Equation 21.25 from t = 0 s to t = 300 s  t  t k(To − Ti ) Qo = q˙ o dt = dt 1/2 0 0 (t) = 2k(To − Ti )(t/) 1/2 = (2)(13.4 W/m-K)(50◦ C − 27◦ C)[(300 s/(370 × 10−6 m/s2 )]1/2 = 3.131 × 105 J/m2

(313.1 kJ/m2 ) ⇐

(b) Here Equation 21.27 is applicable. T(x, t) = Ti +

2q˙ o (t/) 1/2 −x2 /4t q˙ o x e − erfc() k k

and with t (3.70 × 10−6 m2 /s)(300 s) = = 3.533 × 10−4 m2   we obtain T(0.02 m, 300 s) = 27◦ C +

2(104 W/m2 )(3.533 × 10−4 m2 ) 1/2 −(0.30)2 e 13.4 W/m-K −

(104 W/m2 )(0.02 m) (0.6714) 13.4 W/m-K

= 27◦ C + 25.6◦ C − 10.0◦ C = 42.6◦ C ⇐ Because q˙ o is a constant, the cumulative energy supplied over a time, t, is Qo = q˙ o t = (104 W/m2 )(300 s) = 3.0 × 106 J/m2

(3000 kJ/m2 ) ⇐

Water pipes are frequently buried in the ground to prevent their freezing and the minimum depth for a prescribed soil along with the associated temperature conditions is of more than casual interest. A design example that deals with the freezing of buried pipes now follows.

21.4

Design Example 12

Consider a parallel line of water pipes that are buried in a soil having a thermal diffusivity of  = 0.138 × 10−6 m/s2 . The surface of the soil is initially at 10◦ C but suddenly drops to −10◦ C and remains at this temperature for 80 days. Determine the minimum burial depth to prevent the freezing of the pipes.

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Introduction to Thermal and Fluid Engineering

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is negligible heat loss to and from the surroundings. (4) A semi-infinite solid is presumed. (5) The thermal diffusivity of the soil is specified. Here, the time of interest is t = (80 days)(24 h/day)(3600 s/h) = 6.912 × 106 s and we solve for x in Equation 21.22. With T(x, t) = 0◦ C,

To = −10◦ C

and

Ti = 10◦ C

a modification of Equation 21.22 yields T(x, t) − Ti = erfc() To − Ti where =

x 2(t) 1/2

Thus, erfc() =

0◦ C − 10◦ C = 0.5000 −10◦ C − 10◦ C

and Table 21.1 gives (with interpolation)  = 0.4795. The burial depth is x = 2(t) 1/2 = 2(0.4795)[(0.138 × 10−6 m/s2 )(6.912 × 106 s)]1/2 = 2(0.4795)(0.9539 m2 ) 1/2 = 0.937 m ⇐

21.5

Finite-Sized Solids

21.5.1 The Long Plane Wall Figure 21.6 shows a long plane wall of thickness, 2L. The wall, which has a thermal conductivity, k, a density, , a specific heat, c, and a thermal diffusivity, , is initially at a uniform temperature, Ti . At time t = 0 and for all times thereafter, both surfaces of the wall are suddenly exposed to an environment that is characterized by a temperature, T∞ < Ti and a

Unsteady-State Conduction

679 k ρ c α Ti

T∞

T∞

h

h

L

L x=0

FIGURE 21.6 Long, solid plane wall with thickness 2L .

heat transfer coefficient, h. Although we are considering convective cooling at the surfaces, the final results are equally applicable to convective surface heating. The cooling of the wall is governed by the Fourier equation in one dimension. ∂2 T ∂x2

=

1 ∂T  ∂t

(21.18)

The initial and the boundary conditions to be imposed on Equation 21.18 are T(x, 0) = Ti ∂T ∂x

(0, t) = 0

(21.30a)

(21.30b)

and −k

∂T ∂x

(L , t) = h[T(L , t) − T∞ ]

(21.30c)

where Equation 21.30b is a consequence of the thermal symmetry that exists about x = 0 and Equation 21.30c represents the balance between the conductive and convective fluxes at x = L. The solution of Equation 21.18 with the boundary conditions of Equation 21.30 may be obtained using the method of separation of the variables. Because this method is somewhat beyond our scope, we list the solution in terms of a dimensionless temperature ≡

T(x, t) − T∞ Ti − T∞

as =2

∞  n=1

sin n L cos n x 2 e −n t n L + sin n L cos n L

(21.31)

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Introduction to Thermal and Fluid Engineering

where the n ’s are given by the transcendental equation n L tan n L = Bi with Bi = h L/k. The instantaneous rate of heat loss from both faces (x = ±L) may be found from ˙ = −2k A∂T (L , t) Q ∂x or ˙ = 4k A(Ti − T∞ ) Q

∞  n=1

n sin2 n L 2 e −n t n L + sin n L cos n L

(21.32)

and the cumulative heat loss over a period of time, t, from  t ˙ dt Q= Q 0

or Q = 4c A(Ti − T∞ )

∞  n=1

sin2 n L 2 (1 − e −n t ) n (n L + sin n L cos n L)

(21.33)

The evaluation of Equations 21.31, 21.32, and 21.33 involves a laborious computational procedure and we will resort to a graphical method outlined in the paragraphs that follow. It can be shown using Equation 21.31 that the dimensionless temperature, , is a function of X = x/L and the Fourier and Biot moduli Fo ≡

t L2

and

Bi ≡

hL k

Thus,  = f ( X, Fo, Bi) It follows that the dimensionless temperature at x = 0 is given by o =

T(0, t) − T∞ To − T∞ = = f (Fo, Bi) Ti − T∞ Ti − T∞

This relationship is plotted in Figure 21.7a. If the ratio, /o , is formed, the ratio becomes a function of x/L and Bi. The ratio, /o , is plotted as a function of 1/Bi instead of Bi for several values of x/L in Figure 21.7b. With Qi = 2L Ac(Ti − T∞ ) Equation 21.33 can be used to create the ratio, Q/Qi , which is then found to be a function of Fo and Bi. It is noted that Qi is the initial energy content of the wall above the environment and is the maximum energy that the wall can give to the environment as it cools down to the temperature of the environment. Figure 21.7c provides a graph of Q/Qi as a function of FoBi2 for various values of Bi. We note that as t −→ ∞, Fo −→ ∞, FoBi −→ ∞ and Q −→ Qi , or Q/Qi = 1. The graphs in Figure 21.7a and 21.7b were originally developed by Heisler (1947) and are known as the Heisler charts.

Unsteady-State Conduction

681

Bi

10

1 12 4

9

8 25 20 18 6 1

7 6 5

4

.02

3 2.5 2 1.8.6 1 1.4 1.2

.07 .05 .04 .03

0 10 0 9 80 70 60 50 45 40 35 30

.2 .01

1.0 0.8 0.7.6 0 .5 0 .4 0 .3 0 0.2.1 0 .05 0 0

θo/θi = (To – T∞)/(Ti – T∞)

10 .7 .5 .4 .3

.01 .007 .005 .004 .003 .002 .001 0

1

2

3

4

6 8 10 12 14 16 18 20 22 24 26 28 30 40 50 60 70 80 90 100110120130140 150 200 300

400

500

600 700

Fo = αt L2 (a) 1.0 0.9

0 X/L = 0.2

0.8

0.4

θ/θ0 = (T–T∞)/(T0–T∞)

0.7 0.6

0.6

0.5 0.4 0.8 0.3 0.9 0.2 0.1 0 0.01 0.02

1.0

0.05 0.1

0.2

0.5

1.0

2

3 5

10

20

50

100

k 1 = hL Bi (b) FIGURE 21.7 Charts for unsteady conduction in a long, solid plane wall (a) mid-plane temperature, (b) temperature as a function of the mid-plane temperature, and (c) dimensionless heat loss. Figures 21.7a and 21.7b are reproduced from the original Heisler paper [Heisler (1947), courtesy of ASME]. Figure 21.7c is attributed to Groeber ¨ et al. (1961).

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Introduction to Thermal and Fluid Engineering

10–5

10–4

10–3

10–2

10–1

1

10

50

20

hL k

5

Bi =

10

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Bi = 0.00 1 0.00 2 0.00 5 0.01 0.02 0.05 0.1 0.2 0.5 1 2

Q/Qi

1.0 0.9

102

103

104

h2αt/k2 (c) FIGURE 21.7 Continued

Other charts for the rapid solution of unsteady or transient conduction problems were developed by Gurney and Lurie (1925), Groeber ¨ (1925), Boelter et al. (1965), and Schneider (1965). Example 21.4 illustrates the use of Figure 21.7.

Example 21.4 As sketched in Figure 21.8, an AISI 316 stainless steel plate (k = 12 W/m-K,

 = 3.48×10−6 m2 /s,  = 8238 kg/m4 and c = 468 J/kg-K) is 6-cm thick and is originally at a uniform initial temperature of 325◦ C. The plate is cooled by cold air jets at 25◦ C that provide a convective heat transfer coefficient of 400 W/m2 -K. Determine (a) the time required to bring the centerline temperature to 85◦ C, (b) the surface temperature when the centerline temperature is 85◦ C, and (c) the cumulative heat loss to the air per unit area of the plate surface. k = 12 W/m-K α = 3.48 × 10–6 m2/s ρ = 8238 kg/m3 c = +68 J/kg-k Ti = 325°C

h = 400 W/m2-K

T∞ = 25°C

h = 400 W/m2-k

T∞ = 25°C

6 cm

FIGURE 21.8 Configuration for Example 21.4.

Unsteady-State Conduction

683

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (a) We first determine the dimensionless temperature, o and the inverse of the Biot number, Bi−1 o =

T(0, t) − T∞ 85◦ C − 25◦ C = = 0.20 Ti − T∞ 325◦ C − 25◦ C

and with L = 0.06 m/2 = 0.03 m Bi−1 =

k 12 W/m-K = = 1.0 hL (400 W/m2 -K)(0.03 m)

With o and Bi−1 known, Figure 21.7a may be read to find Fo =

t = 2.40 L2

which gives t=

2.40L 2 (2.40)(0.03 m) 2 = = 620.7 s  3.48 × 10−6 m2 /s

(10.35 min) ⇐

(b) With Bi = 1.0 and X = 1.0 at the surface, we read Figure 21.7b to obtain T(L , t) − T∞ = 0.632 T(0, t) − T∞ Thus, the surface temperature, T(L , t) is T(L , t) = T∞ + 0.632[T(0, t) − T∞ ] = 25◦ C + 0.632(85◦ C − 25◦ C) = 25◦ C + 37.9◦ C = 62.9◦ C ⇐ (c) With FoBi2 = (2.4)(1.0) 2 = 2.4 and Bi = 1.0, we may read Figure 21.7c to obtain Q = 0.80 Qi so that Q = (0.80) Qi = (0.80)[2L Ac(Ti − T∞ )] or Q = (0.80)(2)(0.03 m)(8238 kg/m3 )(468 J/kg-K)(325◦ C − 25◦ C) A = 5.552 × 107 J/m2 ⇐

684

Introduction to Thermal and Fluid Engineering TABLE 21.2

Values of C1 and 1 Used in the One-Term Approximation to the Series Solution for Transient One-Dimensional Conduction in the Plane Wall Bi

β1 , rad

C1

Bi

β1 , rad

C1

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.15 0.20 0.25 0.30 0.40 0.50 0.60 0.70

0.0998 0.1410 0.1732 0.1987 0.2217 0.2425 0.2615 0.2791 0.2956 0.3111 0.3779 0.4328 0.4801 0.5218 0.5932 0.6533 0.7051 0.7506

1.0017 1.0033 1.0049 1.0066 1.0082 1.0098 1.0114 1.0130 1.0145 1.0160 1.0237 1.0311 1.0382 1.0450 1.0580 1.0701 1.0814 1.0919

0.80 0.90 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 21.0 30.0 40.0 50.0 100 ∞

0.7910 0.8274 0.8603 1.0769 1.1925 1.2646 1.3138 1.3496 1.3766 1.3978 1.4149 1.4289 1.4961 1.5202 1.5325 1.5400 1.5552 1.5707

1.1016 1.1107 1.1191 1.1725 1.2102 1.2287 1.2402 1.2479 1.2532 1.2570 1.2598 1.2620 1.2699 1.2717 1.2723 1.2727 1.2731 1.2733

21.5.2 One-Term Approximate Solutions When Fo > 0.20, the first term of the series in Equations 21.31 and 21.33 provides sufficiently accurate results. These one-term approximate solutions are =

 x T(x, t) − T∞ 2 = C1 e − 1 Fo cos 1 Ti − T∞ L

(21.34)

o = C1 e − 1 Fo

(21.35)

Q sin 1 =1− o Qi 1

(21.36)

2

and

The values of C1 and 1 for a range of values of Bi are provided in Table 21.2.

Example 21.5 Use the one-term approximate solutions for the long plane wall to develop the solution to Example 21.4.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (4) The one-term approximate solution is to be used.

Unsteady-State Conduction

685

(a) For Bi = 1, Table 21.2 gives 1 = 0.8603 rad and C1 = 1.1191. Solving Equation 21.35 for Fo with o = 0.20 from Example 21.4 gives Fo = −

      1 o 1 C1 1 1.1191 ln = ln = ln C1 o (0.8603) 2 0.2 21 21

= (1.3511) ln 5.5955 = (1.3511)(1.7220) = 2.327 which then gives t=

2.327L 2 2.327(0.03 m) 2 = = 601.8 s  3.48 × 10−6 m2 /s

(10.03 min) ⇐

(b) Using Equation 21.34 with X = x/L = 1 gives for the surface T(L , t) − T∞ 2 = C1 e − 1 Fo cos 1 Ti − T∞ = 1.1191e −(0.8603)

2

(2.327)

cos 0.8603

= 1.1191e −1.7223 (0.6522) = 0.130 and hence, T(L , t) = 0.130(Ti − T∞ ) + T∞ = 0.130(325◦ C − 25◦ C) + 25◦ C = 64◦ C ⇐ (c) Use of Equation 21.36 gives Q sin 1 = 1− o Qi 1 = 1−

sin 0.8603 (0.20) 0.8603

= 1 − (0.8811)(0.20) = 1 − 0.1762 = 0.8238 which then provides Q = (0.8238) Qi = (0.8238)(2L Ac(Ti − T∞ ) or Q = (0.8238)(2)(0.03 m)(8238 kg/m3 )(468 J/kg-K)(325◦ C − 25◦ C) A = 5.72 × 107 J/m2 ⇐ These results compare favorably with those obtained in Example 21.4.

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Introduction to Thermal and Fluid Engineering

h

ro k ρ c α Ti T∞

FIGURE 21.9 Long solid cylinder of radius, ro .

21.5.3 The Long Solid Cylinder Figure 21.9 shows a long solid cylinder of radius, ro , initially at a uniform temperature of Ti . The cylinder has thermal conductivity, k, density, , specific heat, c, and thermal diffusivity, . At time t = 0, the cylinder is placed in an environment at temperature, T∞ < Ti , which cools the lateral surface of the cylinder through a heat transfer coefficient, h. Although convective surface cooling is considered, the results obtained are also applicable for the case of convective surface heating. The partial differential equation describing the cooling of the cylinder is the Fourier equation in one dimension but in cylindrical coordinates ∂2 T ∂r 2

+

1 ∂T 1 ∂T = r ∂r  ∂t

(21.37)

The initial and the boundary conditions to be satisfied are: T(r, 0) = Ti ∂T ∂r

(0, t) = 0

(21.38a)

(21.38b)

and −k

∂T ∂r

(ro , t) = h[T(ro , t) − T∞ ]

(21.38c)

where Equation 21.38b describes the thermal symmetry about the axis of the cylinder (at r = 0) and Equation 21.38c is the result of the surface energy balance at r = ro . The solution of Equation 21.37, which satisfies the boundary conditions of Equation 21.38, may also be written in terms of a dimensionless temperature ≡

T(r, t) − T∞ Ti − T∞

Unsteady-State Conduction

687

and is =2

∞  1 J 1 (nro ) J 0 (nr ) 2 e −n t 2 2  r J ( r ) + J ( r ) n o n o 1 n=1 n o 0

(21.39)

where the n ’s are solutions to the transcendental equation nro

J 1 (nro ) = Bi J 0 (nro )

(21.40)

where Bi = hro /k and J 0 and J 1 are the Bessel functions of the first kind of order 0 and 1, respectively. The instantaneous rate of heat loss from the surface of the cylinder is ˙ Q ∂T = −2ro k (ro , t) L ∂r or ∞  ˙ J 12 (nro ) Q 2 = 4k(Ti − T∞ ) e −n t 2 2 L n=1 J 0 (n r o ) + J 1 (n ro )

(21.41)

and the cumulative heat loss over a period of time, t, will be ∞  J 12 (nro ) Q 2 = 4c(Ti − T∞ ) (1 − e −n t ) 2 2 2 L  [J ( r ) + J ( r )] n o n o 1 n=1 n 0

(21.42)

In this case, the dimensionless temperature is a function of R = r/ro , Fo = t/ro2 and Bi = hro /k, that is  = f ( R, Fo, Bi) Similarly, o =

T(0, t) − T∞ To − T∞ = = f (Fo, Bi) Ti − T∞ Ti − T∞

and Q = f (Fo, Bi) Qi where Qi = ro2 Lc(Ti − T∞ ) is the initial internal energy of the cylinder above the environment. Figure 21.10a shows o as a function of Fo and Bi. The ratio, /o as a function of Bi for different values of R appears in Figure 21.10b and Figure 21.10c provides a graph of Q/Qi as a function of Bi2 Fo for various values of Bi. These charts, like those of Figure 21.7, are also Heisler and Groeber ¨ charts. Example 21.6 illustrates the use of Figure 21.10.

Example 21.6 A cylindrical rod of radius 2 cm is made of a composite material whose properties are k = 10 W/m-K,  = 1300 kg/m3 , and c = 1500J/kg-K. The rod (Figure 21.11), which is initially at a uniform initial temperature of 25◦ C, is suddenly exposed to a

688

Introduction to Thermal and Fluid Engineering

1.0 .7 m =5 4 3.5 3. 2.5 0

.2

m

.5

2.0

1.81.6 .4 1 2 1. .0 1

0.1 .07 .05 .04

00

70

8

60

50 45 40

0

.007

=1

90 80

6

.01

m

7

.02

Bi

10 9

0.8

.03

=2

20 18 16 14 12

0.60.5 0.4 0.3 0.2 0.1

35

θo/θi = (To – T∞)/(Ti – T∞)

.5 .4 .3

30

.005 .004 .003 .002 .001 0

1

2

3

4

6

8

10 12 14 16 18 20 22 24 26 28 30 40 50 60 70 80 90 100 110 120 130 140 150 200

300350

αt r 2o (a)

Fo =

0 n + 0.2

10 0.9

Cylinder 0.4

0.8

θ/θo = (T – To )/(To – T∞)

0.7 0.6

0.8

0.5 0.4 0.8

0.3 0.2

0.8

0.1 1.0 0 .01

.02 .03 .05

.1

.2 .3

.5

1.0 2 3 k 1 = hL Bi

5

10

20 30 50

100

(b) FIGURE 21.10 Charts for unsteady conduction in a long, solid cylinder (a) centerline temperature, (b) temperature as a function of the mid-plane temperature, and (c) dimensionless heat loss. Figures 21.10a and 21.10b are reproduced from the original Heisler paper [Heisler (1947), courtesy of ASME] and Figure 21.10c is attributed to Groeber ¨ et al. (1961).

Unsteady-State Conduction

689

50

20

10

5

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Bi = 0.00 1 0.00 2 0.00 5 0.01 0.02 0.05 0.1 0.2 0.5 1 2

Q/Qi

1.0 0.9

hr Bi = k 0 10–5

10–4

10–3

10–2

10–1

1

10

102

103

104

h2αt/k2 (c) FIGURE 21.10 Continued

stream of hot exhaust gases at 125◦ C that provides a heat transfer coefficient of 250 W/m2 -K. Determine (a) the temperature at the center of the cylinder 2 min after exposure to the exhaust gases, (b) the temperature at a radial distance of 1.2 cm from the center at 2 min, and (c) the heat gained by the cylinder in 2 min.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. h = 250 W/m2-K

k = 10 W/m-K ρ = 1300 kg/m3 c = 1500 J/kg-K Ti = 25°C

T∞ = 125°C

ro

FIGURE 21.11 Configuration for Example 21.6.

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Introduction to Thermal and Fluid Engineering

(a) We first determine Fo and Bi−1 . With t = (2 min)(60 s/min) = 120 s Fo = =

t kt = ro2 cro2 (10 W/m-K)(120 s) (1300 kg/m3 )(1500 kJ/kg-K)(0.02 m) 2

= 1.54 and Bi−1 =

k 10 W/m-K = = 2.0 hro (250 W/m2 -K)(0.02 m)

With Fo and Bi−1 known, Figure 21.10a may be read to find o =

T(0, t) − T∞ = 0.29 Ti − T∞

which then gives T(0, t) = (0.29)(Ti − T∞ ) + T∞ = (0.29)(25◦ C − 125◦ C) + 125◦ C = −29◦ C + 125◦ C = 96◦ C ⇐ (b) At a radial distance of 1.2 cm R=

r 1.2 cm = = 0.60 ro 2 cm

and with Bi−1 = 2, Figure 21.10b reveals that T(r, t) − T∞ = 0.92 T(0, t) − T∞ so that T(1.2 cm, 120 s) = 0.92(96◦ C − 125◦ C) + 125◦ C = 98.3◦ C ⇐ (c) With FoBi2 = (1.54)(0.50) 2 = 0.385 and Bi = 0.5, we may read Figure 21.10c to obtain Q = 0.75 Qi so that Q = (0.75) Qi = (0.75)ro2 Lc(Ti − T∞ ) or Q = (0.75)(0.02 m) 2 (1300 kg/m3 )(1500 J/kg-K(25◦ C − 125◦ C) L = −1.837 × 105 J/m ⇐ The heat gained by the rod is 183.7 kJ per meter of rod length.

Unsteady-State Conduction

691

TABLE 21.3

Values of C1 and 1 Used in the One-Term Approximation to the Series Solution for Transient One-Dimensional Conduction in the Long Cylinder Bi

β1 , rad

C1

Bi

β1 , rad

C1

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.15 0.20 0.25 0.30 0.40 0.50 0.60 0.70

0.1412 0.1995 0.2439 0.2814 0.3142 0.3438 0.3708 0.3960 0.4195 0.4417 0.5376 0.6170 0.6856 0.7465 0.8516 0.9408 1.0185 1.0873

1.0025 1.0050 1.0075 1.0099 1.0124 1.0148 1.0173 1.0197 1.0222 1.0246 1.0365 1.0483 1.0598 1.0712 1.0932 1.1143 1.1346 1.1539

0.80 0.90 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 21.0 30.0 40.0 50.0 100 ∞

1.1490 1.2048 1.2558 1.5995 1.7887 1.9081 1.9898 2.0490 2.0937 2.1286 2.1566 2.1795 2.2881 2.3261 2.3455 2.3572 2.3809 2.4050

1.1725 1.1902 1.2071 1.3384 1.4191 1.4698 1.5029 1.5253 1.5411 1.5526 1.5611 1.5677 1.5919 1.5973 1.5993 1.6002 1.6015 1.6018

21.5.4 One-Term Approximate Solutions For Fo > 0.20, the one-term approximate solutions are =

T(r, t) − T∞ 2 = C1 e − 1 Fo J 0 ( 1 R) Ti − T∞

(21.43)

T(0, t) − T∞ 2 = C1 e − 1 Fo Ti − T∞

(21.44)

Q 2o =1− J 1 ( 1 ) Qi 1

(21.45)

here R = r/ro . o = and

where 1 = 1ro . The values of C1 and 1 for a range of values of Bi are provided in Table 21.3 and Table 21.4 provides values of J 0 (x) and J 1 (x), which are needed when using Equations 21.43 and 21.45.

Example 21.7 Use the one-term approximate solutions for the long cylinder to develop the solution to Example 21.6.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (4) The one-term approximate solution is to be used.

692

Introduction to Thermal and Fluid Engineering TABLE 21.4

Values of the Bessel Functions, J 0 (x) and J 1 (x) x

J0 (x)

J1 (x)

x

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00

1.0000 0.9975 0.9900 0.9776 0.9604 0.9385 0.9120 0.8812 0.8463 0.8075 0.7652 0.7196 0.6711 0.6201 0.5669 0.5118 0.4554 0.3980 0.3400 0.2818 0.2239

0.0000 0.0499 0.0995 0.1483 0.1960 0.2423 0.2867 0.3290 0.3688 0.4060 0.4401 0.4709 0.4983 0.5220 0.5420 0.5579 0.5699 0.5778 0.5815 0.5812 0.5767

2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 3.20 3.40 3.60 3.80 4.00 4.20 4.40 4.60 4.80 5.00

J0 (x)

J1 (x)

0.1666 0.1104 0.0554 0.0025 −0.0484 −0.0968 −0.1425 −0.1850 −0.2243 −0.2601 −0.3202 −0.3643 −0.3918 −0.4026 −0.3972 −0.3766 −0.3423 −0.2961 −0.2404 −0.1776

0.5683 0.5560 0.5399 0.5202 0.4971 0.4708 0.4416 0.4097 0.3754 0.3391 0.2613 0.1792 0.0955 0.0128 −0.0661 −0.1387 −0.2028 −0.2566 −0.2985 −0.3276

(a) For Bi = 0.5, Table 21.3 gives for the cylinder, 1 = 0.9408 rad and C1 = 1.1143. Solving Equation 21.44 for o with Fo = 1.54 gives o = C1 e − 1 Fo = (1.1143)e −(0.9408) 2

2

(1.54)

= (1.1143)e −1.3631 = (1.1143)(0.2559) = 0.2851 which then gives T(0, t) = (0.2851)(Ti − T∞ ) + T∞ = (0.2851)(25◦ C − 125◦ C) + 125◦ C = −28.5◦ C + 125◦ C = 96.5◦ C ⇐ (b) With R = r/ro = 0.6 and 1 = 0.9408 rad 1 R = (0.9408)(0.6) = 0.5645 and, with interpolation, Table 21.4 provides J 0 (0.5645) = 0.9214 Then  = C1 e − 1 Fo J 0 ( 1 R) 2

= (1.1143)e −(0.9408)

2

(1.54)

0.9214)

= (1.1143)(0.2559)(0.9214) = 0.2627

Unsteady-State Conduction

693

Thus, T(1.2 cm, 120 s) = (0.263)(Ti − T∞ ) + T∞ = (0.263)(25◦ C − 125◦ C) + 125◦ C = −26.3◦ C + 125◦ C = 98.7◦ C ⇐ (c) Table 21.4 shows that J 1 ( 1 ) = J 1 (0.9408) = 0.4199 (with interpolation). Equation 21.45 then gives Q 2o = 1− J 1 ( 1 ) Qi 1 = 1−

(2)(0.2851) (0.4199) 0.9408

= 0.746 and with Q = 245,000 J/m Q = (0.746) Qi = (0.746)(−245, 000 J/min) = −1.828 × 105 J/min

21.5.5 The Solid Sphere Figure 21.12 shows a solid sphere of radius, ro made of a material with thermal conductivity, k, density, , specific heat, c, and thermal diffusivity, . The sphere is initially at a uniform temperature, Ti . At time, t = 0, the sphere is exposed to an environment at temperature T∞ , which is less than Ti . The convective heat transfer coefficient between the surface of the sphere and the environment is h. Although convective cooling is considered, the final results are equally applicable to convective heating.

h k c ρ α Ti T∞

FIGURE 21.12 Solid sphere of radius, ro .

694

Introduction to Thermal and Fluid Engineering

The cooling of the sphere is described by the Fourier equation in the radial direction but in spherical coordinates ∂2 T ∂r 2

+

2 ∂T 1 ∂T = r ∂r  ∂t

(21.46)

Here Equation 21.38a provides the initial condition T(r, 0) = Ti

(21.38a)

and Equations 21.38b and 21.38c provide the boundary conditions ∂T ∂r

(0, t) = 0

(21.38b)

and ∂T

−k

∂r

(ro , t) = h[T(ro , t) − T∞ ]

(21.38c)

so that the solution to Equation 21.46 is given by =

∞  T(r, t) − T∞ sin nro − nro cos nro sin nr −2n t =2 e Ti − T∞  r − sin nro cos nro nr n=1 n o

(21.47)

where the n s are the positive roots of tan nro −

nro =0 1 − Bi

(n = 1, 2, 3, · · · , ∞)

(21.48)

and Bi =

hro k

The instantaneous heat loss from the surface of the sphere is ˙ = 8kro (Ti − T∞ ) Q =

∞  n=1

(sin nro − nro cosnro ) 2 −2n t e nro (nro − sin nro cos nro )

(21.49)

and the cumulative heat loss over a period of time, t, is  t ˙ dt Q= Q 0

or Q = 8c(Ti − T∞ ) ∞   (sin nro − nro cos nro ) 2  2 1 − e −n t 3  ( r − sin nro cos nro ) n=1 n n o

(21.50)

As in the case of a cylinder,  may be expressed as a function of R = r/ro , Fo = t/ro and Bi = hro /k, that is,  = f ( R, Fo, Bi). Likewise o = f (Fo, Bi) and Q/Qi = f (Fo, Bi) where Qi =

4 3 r c(Ti − T∞ ) 3 o

Unsteady-State Conduction

695

is the initial internal energy of the solid above the environment. Figure 21.13a shows o as a function of Fo and 1/Bi. The ratio /o as a function of 1/Bi for various values of R is plotted in Figure 21.13b. The ratio Q/Qi is plotted as a function of Bi2 Fo for various values of Bi in Figure 21.13c.

Example 21.8 In an experimental set-up, a 4-cm-diameter copper ball (k = 401 W/m-K, c = 385J/kg-K, and  = 8933 kg/m3 ) is heated in a furnace to a uniform temperature of 100◦ C and suddenly immersed in water at 25◦ C. A thermocouple placed at the center of the ball records a temperature of 44.5◦ C after 2 min of immersion. Use the appropriate Heisler chart to determine the convective heat transfer coefficient on the outer surface of the ball.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. We first calculate o and Fo corresponding to 2 min or 120 s. o =

T(0, 120 s) − T∞ 44.5◦ C − 25◦ C = = 0.26 Ti − T∞ 100◦ C − 25◦ C

and Fo =

t kt = 2 ro cro2

so that Fo =

(401 W/m-K)(120 s) (8933 kg/m3 )(385 J/kg-K)(0.02 m) 2

= 34.98

With o = 0.26 and Fo = 34.98, we read Figure 21.13a to obtain 1 k = = 80 Bi hro which then gives h=

k 401 W/m-K = = 250.6 W/m2 -K ⇐ Biro 80(0.02 m)

21.5.6 One-Term Approximate Solutions When Fo exceeds 0.20, the one-term approximate solutions are =

T(r, t) − T∞ sin( 1r/ro ) 2 = C1 e − 1 Fo Ti − T∞ 1r/ro

(21.51)

T(0, t) − T∞ 2 = C1 e − 1 Fo Ti − T∞

(21.52)

o =

696

Introduction to Thermal and Fluid Engineering 1 .7 .5 .4 .3

Bi

m=

m m

0.1 .07

35

4 50 40 5

25

7

208 1 16

6

4 5 3.

1.4 1.2

1.0

.02

9

5

1.6

.05 .04 .03

8

30

=3 2 .0 2 .8 2 .6 2.2 .4 2.0 1.8

=1 4 12

10

10 9 0 80 0 70 60

5 0.7

.01

0.5 0.2 0.1 0.05

.007 .005 .004 .003

0.35

θo/θi = (To – T∞)/(Ti – T∞)

.2

(a)

0

.002 .001 0

.5

1.0

1.5

2

2.5

3

4

5

6

7

8

9 10 15 20 25 30 35 40 45 50 70 90 110 130 150 170 190 210 230 250

Fo =

αt r 2o

(a) 0

1.0

0.02 0.9 0.8

Sphere

0.4

θ/θo = (T – To)/(To – T∞)

0.7 0.6 0.6 0.5 0.4 0.3

0.8

0.2

0.9

0.1 1.0 0 .01

.02 .03 .05.07 .1

.2 .3

.5 .7 1.0

2 3

5 7 10

20 30 50 70 100

k 1 = hL Bi (b) FIGURE 21.13 Charts for unsteady conduction in a solid sphere (a) center temperature, (b) temperature as a function of the center temperature, and (c) dimensionless heat loss. Figures 21.13a and 21.13b are reproduced from the original Heisler paper [Heisler (1947), courtesy of ASME] and Figure 21.13c is attributed to Groeber ¨ et al. (1961).

50

20

10

5

1

2

0.5

0.05 0.1 0.2

0.00 1 0.00 2 0.00 5 0.01 0.02

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

697

Bi =

Q/Qi

Unsteady-State Conduction

Bi =

10–5

10–4

10–3

10–2

10–1

1

10

102

hr0 k 103

104

h2αt/k2 (c) FIGURE 21.13 Continued

and Q 3o = 1 − 3 (sin 1 − 1 cos 1 ) Qi 1

(21.53)

where 1 = 1ro . The values of C1 and 1 for a range of values of Bi are provided in Table 21.5.

Example 21.9 A 4-cm-diameter lead ball (k = 35.3 W/m-K, c = 129 J/kg-K, and  =

11,433 kg/m3 ) at 300◦ C as in Figure 21.14, is suddenly placed in a water jet. At 30◦ C, the water jet produces a heat transfer coefficient of 1765 W/m2 -K. Use the one-term approximate solutions, determine (a) the center and the surface temperatures after ten seconds, TABLE 21.5

Values of C1 and 1 Used in the One-Term Approximation to the Series Solution for Transient One-Dimensional Conduction in the Solid Sphere Bi

β1 , rad

C1

Bi

β1 , rad

C1

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.15 0.20 0.25 0.30 0.40 0.50 0.60 0.70

0.1730 0.2445 0.2989 0.3450 0.3852 0.4217 0.4550 0.4860 0.5150 0.5423 0.6608 0.7593 0.8448 0.9208 1.0528 1.1656 1.2644 1.3525

1.0030 1.0060 1.0090 1.0120 1.0149 1.0179 1.0209 1.0239 1.0268 1.0298 1.0445 1.0592 1.0737 1.0880 1.1164 1.1441 1.1713 1.1978

0.80 0.90 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 21.0 30.0 40.0 50.0 100 ∞

1.4320 1.5044 1.5708 2.0288 2.2889 2.4456 2.5704 2.6537 2.7165 2.7654 2.8044 2.8363 2.9857 3.0372 3.0632 3.0788 3.1102 3.1415

1.2236 1.2488 1.2732 1.4793 1.6227 1.7201 1.7870 1.8338 1.8674 1.8921 1.9106 1.9249 1.9781 1.9898 1.9942 1.9962 1.9990 2.0000

698

Introduction to Thermal and Fluid Engineering k = 35.3 W/m-K c = 129 J/kg-K ρ = 11,433 kg/m3 h = 1765 W/m2-K

4 cm

T∞ = 30°C

FIGURE 21.14 Configuration for Example 21.9.

(b) the initial energy of the ball above the water jet environment, and (c) the cumulative heat loss over a period of 10 s.

Solution Assumptions and Specifications (1) Unsteady-state conditions exist. (2) Thermal properties do not vary with temperature. (3) There is no radiation exchange with the surroundings. (4) The one-term approximate solution is to be used. (a) We first calculate Bi and Fo. Bi =

hro (1765 W/m2 -K)(0.02 cm) = = 1.0 k 35.3 W/m-K

and Fo =

kt (35.3 W/m-K)(10 s) = = 0.60 2 cro (11, 433 kg/m3 )(129 J/kg-K)(0.02 m) 2

With Bi = 1.0, Table 21.5 gives for the sphere, 1 = 1.5708 rad and C1 = 1.2732. Solving Equation 21.52 for o gives o = C1 e − 1 Fo 2

= (1.2732)e −(1.5708)

2

(0.60)

= (1.2732)e −1.4804 = (1.2732)(0.2275) = 0.2897 which then gives T(0, 10 s) = (0.2897)(Ti − T∞ ) + T∞ = (0.2897)(300◦ C − 30◦ C) + 30◦ C = 78.2◦ C + 30◦ C = 108.2◦ C ⇐

Unsteady-State Conduction

699

The surface temperature may be obtained from Equation 21.51 T(r, t) − T∞ sin( 1r/ro ) = o Ti − T∞ 1r/ro = (0.2897)

sin 1.5708 = 0.1844 1.5708

which gives T(0.02 m, 10 s) = (0.1844)(Ti − T∞ ) + T∞ = (0.1844)(300◦ C − 30◦ C) + 30◦ C = 49.8◦ C + 30◦ C = 79.8◦ C ⇐ (b) The initial energy of the ball above the water jet environment may be found as Qi = =

4ro3 c(Ti − T∞ ) 3 4(0.02 m) 3 (11,433 kg/m3 )(129 J/kg-K)(300◦ C − 30◦ C) 3

= 13,344 J ⇐ Using Equation 21.53, we have  3o Q = Qi 1 − 3 (sin 1 − 1 cos 1 ) 1 

3(0.2897) = (13,344 J) 1 − (sin 1.5708 − 1.5708 cos 1.5708) (1.5708) 3



= 10,352 J

21.6

Summary

Certain transient heating or cooling problems can be handled by the lumped capacitance model in which the Biot modulus is less than 0.10 Bi =

h Lc < 0.1 k

where L c is the ratio of the volume of the solid to its surface area, L c = V/S. In this case, from Equation 21.5, we have T − T∞ = e t/ Ti − T∞

(21.5)

where the time constant, , is defined by  = Vc/hS

(21.6)

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Introduction to Thermal and Fluid Engineering

In the event that internal heat generation, E˙ g , is present, the temperature-time history is given by E˙ g − hS(T − T∞ ) = e −t/ E˙ g − hS(Ti − T∞ )

(21.16)

The semi-infinite model considers the solid as extending to infinity in all dimensions except for one surface. For a constant surface temperature, To T(x, t) = Ti + (To − Ti ) erfc()

(21.22)

where erfc is the complementary error function tabulated in Table 21.1 and where =

x 2(t) 1/2

(21.23)

The surface heat flux for a constant surface temperature is given by q˙ o =

k(To − Ti ) (t) 1/2

(21.25)

The temperature for constant heat flux is given by Equation 21.27. For surface convection where the convection is specified, Equation 21.29 pertains. The Heisler and Groeber ¨ charts provide approximations to the unsteady-heat flow with convection present and are reproduced for the long, solid plane wall in Figure 21.7. They are based on a dimensionless spatial coordinate, X = x/L, and the Fourier and Biot moduli Fo ≡

t L2

and

Bi ≡

hL k

The charts give  = f ( X, Fo, Bi),

o = f (Fo, Bi),

and

Q = f (Fo, Bi) Qi

For the long solid cylinder (Figure 21.10) and the solid sphere (Figure 21.13), R = r/ro Fo ≡

t ro2

and

Bi ≡

hro k

and the charts give  = f ( X, Fo, Bi),

o = f (Fo, Bi),

and

Q = f (Fo, Bi) Qi

The classical solutions for the long, solid plane wall, the long solid cylinder and the solid sphere involve a series summation with each term in the series based upon a particular eigenvalue. One-term approximate solutions are obtainable and are given for , o , and ˙ Q ˙ i for cases where Fo exceeds 0.20. For the long, solid plane wall, Equations 21.34, Q/ 21.35, and 21.36 are applicable. For the long solid cylinder, Equations 21.43, 21.44, and 21.45 pertain; and for the solid sphere, the equations are Equations 21.51, 21.52, and 21.53.

Unsteady-State Conduction

21.7

701

Problems

The Lumped Capacitance Model 21.1: A plain carbon steel cube (k = 60.5 W/m-K) 3 cm on a side is heated to 200◦ C and then allowed to cool in a stream of air at 20◦ C, which provides a convection coefficient of 50 W/m2 -K. Determine the Biot number and check to see whether the lumped capacitance model is valid. 21.2: Reconsider the cube in Problem 21.1. The cube has a density of 7833 kg/m3 and a specific heat of 465J/kg-K. Determine (a) the time needed to cool the cube to 25◦ C and (b) the cumulative heat loss from the cube over this period of time. 21.3: In a laboratory experiment, the forced convection heat transfer coefficient is measured by heating a thermocouple equipped copper sphere with a Bunsen burner and then suddenly exposing the sphere to a stream of air. One such set-up uses a 25.4cm-diameter copper sphere (k = 401 W/m-K and c = 385J/kg-K), which is heated to 320◦ C and allowed to cool in the air at 25◦ C. After 1 min of exposure to the air, the thermocouple records a temperature of 280◦ C. Determine the average heat transfer coefficient between the sphere and the airstream. 21.4: Consider the lumped capacitance cooling of a solid whose specific heat changes linearly with temperature according to c = c ∞ [1 + (T − T∞ )] where c ∞ is the specific heat at T∞ and is a constant. Show that the temperaturetime history of the solid is given by ln

T − T∞ h St + (T − T∞ ) = − Ti − T∞ c ∞

21.5: A steel ball 20 mm in diameter with an initial temperature of 800◦ C is drawn from a furnace. At time t = 0 it is suddenly placed in an oil bath at 20◦ C, which provides a heat transfer coefficient of 120 W/m2 -K. Assuming that the density of steel is  = 8055 kg/m3 and that c = 500[1 + 3 × 10−4 (T − 20◦ C)] with c in J/kg-K and T in ◦ C, determine the time required for the ball to cool down to 300◦ C. 21.6: A 2-cm-thick AISI 302 stainless steel plate initially at a uniform temperature of 520◦ C is suddenly placed in a coolant bath at 25◦ C, which provides a convective heat transfer coefficient of 95 W/m2 -K. Assuming that c = 480J/kg-K,  = 8055 kg/m3 , and k = 15.1 W/m-K, for the plate, determine the Biot number to see if the lumped capacitance model is applicable and, if applicable, determine the time needed for the plate to cool down to 50◦ C. 21.7: Perform the lumped capacitance analysis of Section 21.2.1 to develop an expression for the temperature-time history of the solid when the heat transfer coefficient is temperature dependent in the form   T − T∞ n h = hi Ti − T∞ where h i and n are constants.

702

Introduction to Thermal and Fluid Engineering

21.8: A 2-cm-diameter aluminum ball (c = 903J/kg-K,  = 2702 kg/m3 , and k = 237W/ m-K) initially at a uniform temperature of 90◦ C is suddenly exposed to the air at 25◦ C, which provides a heat transfer coefficient h = 1.60(T − T∞ ) 1/4 W/m2 -K where T is the ball temperature at any instant of time. Assuming that the lumped capacitance model is applicable, determine the temperature of the ball 5 min after exposure to the air. The Effect of Internal Heat Generation 21.9: An electronic component mounted on a heat sink is cooled by the air at 20◦ C, which provides a heat transfer coefficient of 80 W/m2 -K. When the component is turned on, it generates 150 W. Assume the entire assembly to be a lumped capacitance system with a surface area S = 96 cm2 and VC = 50 J/K. Given an initial temperature for the assembly of 20◦ C, determine (a) the temperature of the assembly after 4 min of operation and (b) the steady-state temperature of the assembly. 21.10: The initiation of a chemical reaction in a spherical vessel of 25-cm diameter gives rise to a heat generation of 800 W. The outer surface of the vessel is exposed to the air at 27◦ C, which provides a heat transfer coefficient of 15 W/m2 -K. The vessel and its contents may be regarded as a lumped capacitance system with  = 900 kg/m3 and c = 2600J/kg-K with an initial temperature of 27◦ C. Determine (a) the temperature of the vessel and its contents after 5 min of reaction and (b) the steady-state temperature of the vessel and its contents. 21.11: A 3-cm-diameter thinly coated steel ball heated electrically generates 80 W. When the power is turned on, the ball loses heat to its surroundings by convection and radiation. The convective environment is at 27◦ C, which provides a heat transfer coefficient of 8 W/m2 -K while the emissivity of the surface of the coating is = 0.85. Assuming that the ball behaves as a spatially isothermal solid with  = 8933 kg/m3 and c = 385J/kg-K, determine the temperature of the ball at the instant where dT/dr = 25 K/s. 21.12: An electronic component with Vc = 75 J/K generates 10 W when energized. The component, which is initially at 25◦ C, is cooled by the air at 25◦ C. The temperature of the component reaches 32◦ C after 1 min of operation. The surface area of the component is S = 40 cm2 and a lumped capacitance system may be assumed applicable. Determine the heat transfer coefficient provided by the air. 21.13: In Problem 21.12, how long will it take for the component to reach a temperature of 85◦ C? 21.14: In Problem 21.12, assume that the component is turned off after 4 min. Determine the time required for the temperature to fall back to 40◦ C. 21.15: In Problem 21.12, assume that the heat transfer coefficient provided on the exterior of the component is 80 W/m2 -K. Determine (a) the temperature after 3 min and (b) the steady-state temperature. 21.16: A steel plate that is 12-cm thick (c = 436J/kg-K,  = 7775 kg/m3 , and k = 48 W/m-K) at 627◦ C is plunged into an oil bath at 27◦ C. The heat transfer coefficient at the surface is 420 W/m2 -K. Neglecting the edge effects, determine how long it will take for the plate to cool to 97◦ C.

Unsteady-State Conduction

703

The Semi-Infinite Solid 21.17: The surface at x = 0 cm of a semi-infinite solid initially at 85◦ C throughout is suddenly lowered to 30◦ C and kept at that value. A thermocouple located at x = 25 cm records a temperature of 53.56◦ C after 10 min. Determine the thermal diffusivity of the solid. 21.18: A plane carbon steel plate is initially at a uniform temperature of 27◦ C. At t = 0, the temperature at the surface (x = 0 cm) is suddenly increased to 100◦ C and maintained at that value. The thermal conductivity and thermal diffusivity of the plate are 50.5 W/m-K and 1.70 × 10−6 m/s2 , respectively. Treat the plate as a semiinfinite solid and determine at time, t = 60 s (a) the temperature at x = 5 cm and (b) the surface heat flux. 21.19: A thick piece of material (c = 900J/kg-K,  = 1800 kg/m3 , and k = 3.2 W/m-K) is initially at a surface temperature of 60◦ C. The surface at x = 0 cm is suddenly reduced to 25◦ C and maitained at that value. Plot the temperature and heat flux as a function of time at x = 5 cm. 21.20: A thick tungsten block (c = 132J/kg-K,  = 19, 300 kg/m3 , and k = 174 W/m-K) is initially at a surface temperature of 60◦ C. The surface at x = 0 cm is suddenly exposed to a heat source that creates a constant heat flux of 800 W/m2 at the surface. Determine (a) the temperature at a depth of 1 m after 50 min, (b) the surface temperature after 50 min, and (c) the cumulative heat supply over the 50-min period. 21.21: The thermal diffusivity of a material (k = 75 W/m-K) is determined by imposing a known heat flux at the surface (x = 0 cm) of a thick piece of the material and recording the temperature at a certain depth from the surface. In an experiment, a thermocouple registers a temperature of 54.4◦ C at x = 5 cm and t = 2 min when a heat flux of 5 kW/m2 is established at the surface. Assuming the material to be at 50◦ C initially, determine the thermal diffusivity of the material. 21.22: Consider a thick slab of aluminum (c = 903J/kg-K,  = 2702 kg/m3 and k = 237 W/m-K) which is initially at a uniform temperature of 20◦ C. When the surface at x = 0 cm is suddenly exposed to a constant radiant heat flux, and after 2 min, the temperature at a depth of 12 cm is measured at 44.8◦ C. Determine the radiant heat flux imposed at the surface. 21.23: A concrete wall (c = 880J/kg-K,  = 2100 kg/m3 , and k = 1.4 W/m-K) that is 30-mm thick is initially at a uniform temperature of 22◦ C. At time t = 0, the surface at x = 0 cm is subjected to a constant heat flux of 950 W/m2 . Assuming the wall to behave as a semi-infinite solid, determine the temperature at the two faces of the wall after 2 h of exposure to the heat flux. 21.24: Solve Problem 21.23 if, instead of a constant heat flux, the surface at x = 0 is suddenly exposed to a stream of hot fluid, which promotes a heat transfer coefficient of 80 W/m2 -K. 21.25: A hot, thick slab of steel (c = 477J/kg-K,  = 7000 kg/m3 , and k = 14.9 W/m-K) is suddenly exposed at its surface (x = 0 cm) to a cold stream of water at 25◦ C, which provides a heat transfer coefficient of 240 W/m2 -K. The slab is assumed to be of infinite extent and a thermocouple installed at a depth of 3 cm from the surface records 48◦ C after 3 min. Determine the initial temperature of the slab. 21.26: A thick slab of material (c = 850J/kg-K,  = 2000 kg/m3 , and k = 2 W/m-K) is at uniform temperature of 20◦ C. The surface at x = 0 cm is suddenly brought into contact with a convective environment at a temperature of 90◦ C with a heat transfer

704

Introduction to Thermal and Fluid Engineering coefficient of 200 W/m2 -K. Determine the surface heat flux after 5 min of exposure to the convective environment.

21.27: A thick wooden wall with c = 1255J/kg-K, k = 0.16 W/m-K, and  = 720 kg/m3 , initially at a uniform temperature of 22◦ C and is suddenly exposed at its surface at x = 0 cm to hot gases, which provide a heat transfer coefficient of 40 W/m2 -K. The exposed surface of the wall reaches a temperature of 180◦ C in 10 min. Determine the temperature of the hot gases. 21.28: A very thick concrete slab at 147◦ C is sprayed with a large quantity of water at 27◦ C. Determine how long it will take for a point 5 cm below the surface of the concrete to cool to 62◦ C. The heat transfer coefficient at the concrete surface is 270 W/m2 -K. Finite-Sized Solids 21.29: An aluminum plate with c = 903J/kg-K,  = 2702 kg/m3 , and k = 237 W/m-K is 30-cm thick and is initially at a uniform temperature of 220◦ C. At time t = 0, the plate is suddenly placed in a water stream at 25◦ C, which promotes a heat transfer coefficient of 1000 W/m2 -K on both faces of the plate. Determine (a) the time required to cool the centerline of the plate to 80◦ C, (b) the surface temperature at the time determined in (a), and (c) the cumulative heat loss per unit area to the water over this time period. 21.30: A plastic slab (c = 1480J/kg-K,  = 1150 kg/m3 , and k = 0.28 W/m-K) is 3-cm thick and is initially at a uniform temperature of 75◦ C. At time t = 0, one surface of the slab is suddenly exposed to a cold stream of air at 20◦ C providing a heat transfer coefficient of 20 W/m2 -K. The other surface of the slab is insulated. Determine (a) the temperature of the cooled surface and (b) the temperature of the insulated surface after 45-min exposure to the cold air. 21.31: A 4-cm-thick bronze plate (c = 343J/kg-K,  = 8666 kg/m3 , and k = 26 W/m-K), which is initially at a uniform temperature of 20◦ C, is placed in an oven at 500◦ C that provides a heat transfer coefficient of 108 W/m2 -K on both faces of the plate. If a centerline temperature of 400◦ C is to be obtained, determine (a) how long the plate should stay in the oven and (b) how much energy in kilojoules is gained by the plate during this time. 21.32: For the plastic slab of Problem 21.30 determine the time for the slab to lose half of its original energy to the air. 21.33: A 10-cm-diameter stainless steel rod (c = 480J/kg-K,  = 8055 kg/m3 , and k = 15.1 W/m-K), initially at a uniform temperature of 800◦ C, is removed from a furnace and placed in an oil bath at 25◦ C at time t = 0. The rod loses heat by convection to the oil via a heat transfer coefficient of 55 W/m2 -K. Determine (a) the temperature at the center of the rod after 30 min, (b) the surface temperature after 30 min, and (c) the cumulative heat loss from the rod over 30 min. 21.34: A 8-cm-diameter stainless steel rod (k = 10 W/m-K and  = 4.2 × 10−5 m3 /s) is initially at a uniform temperature of 200◦ C and is to be cooled convectively by a coolant at 25◦ C such that the center of the rod must reach 50◦ C in 5 min. Determine the heat transfer coefficient that the coolant must provide. 21.35: A long steel shaft (k = 40 W/m-K and  = 1.2 × 10−5 m3 /s) of 5 cm radius, initially at a uniform temperature of 400◦ C, is removed from a furnace and placed in a coolant bath at 20◦ C. The heat transfer coefficient is 270 W/m2 -K. Determine the time required for a radial location at 3 cm to attain a temperature of 200◦ C.

Unsteady-State Conduction

705

21.36: A long bronze rod (c = 343J/kg-K,  = 8666 kg/m3 , and k = 25 W/m-K) of 6-cm diameter is initially at a uniform temperature of 300◦ C when it is suddenly placed in a coolant bath at 30◦ C. The convective heat transfer coefficient is 87 W/m2 -K. Determine (a) the time for the rod to lose 60 % of its initial energy to the coolant, (b) the energy lost during the time period found in (a), and (c) the temperature in the center of the rod at the time found in (a). 21.37: A solid stainless steel sphere (k = 15.1 W/m-K and  = 3.91 × 10−6 m3 /s) has a radius of 4 cm. From an initial uniform temperature of 250◦ C, it is suddenly exposed to the air at 25◦ C with a heat transfer coefficient of 20 W/m2 -K. Determine (a) the center temperature, (b) the temperature at a radius of 2.4 cm, and (c) the surface temperature after 30 min of exposure to the air. 21.38: A 12-cm-diameter sphere (k = 15 W/m-K and  = 3.8 × 10−6 m3 /s) is heated in a furnace to a uniform temperature of 80◦ C and then is suddenly placed in flowing water at 20◦ C. A thermocouple located in the center of the sphere records a temperature of 50◦ C after 5 min of exposure to the water. Determine the convective heat transfer coefficient created by the flowing water. 21.39: A 20-cm-diameter spherical pellet of lead (k = 35.3 W/m-K and  = 2.41×10−5 m3 /s) is heated to a uniform temperature of 400◦ C before it is suddenly allowed to quench in a bath of oil at 25◦ C. The convection heat transfer coefficient is 160 W/m2 -K. Determine (a) the time that it will take for the pellet to lose 40% of its energy to the oil and, for this time determine (b) the center temperature, (c) the temperature at a radius of 6 cm, and (d) the surface temperature. One-Term Approximations 21.40: Use the one-term approximation for the long plane wall to solve Problem 21.29. 21.41: Use the one-term approximation for the long plane wall to solve Problem 21.30. 21.42: Use the one-term approximation for the long plane wall to solve Problem 21.31. 21.43: Calculate the Fourier number, Fo = t/ro2 , for the rod in problem 21.33. If Fo > 0.2, use the one-term approximation for the long cylinder in a trial-and-error procedure to solve Problem 21.33. 21.44: Use the one-term approximation for the long cylinder to solve Problem 21.34. 21.45: Use the one-term approximation for the long cylinder to solve Problem 21.35. 21.46: Use the one-term approximation for the long cylinder to solve Problem 21.36. 21.47: Use the one-term approximation for the solid sphere to solve Problem 21.37. 21.48: A 12-cm-diameter sphere (k = 15 W/m-K and  = 3.8 × 10−6 m/s2 ) is heated in a furnace to a uniform temperature of 80◦ C and then suddenly placed in flowing water at 20◦ C. The heat transfer coefficient between the sphere and the water is 60 W/m2 -K and a thermocouple located in the center of the sphere records a temperature of 50◦ C. Using the one-term approximation, determine the time required for the sphere to obtain this temperature. 21.49: Suppose that the lead pellet considered in Problem 21.39 loses 50% of its energy to the oil after immersion at t = 0. Use the one-term approximation to determine the time required for this energy loss.

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22 Forced Convection—Internal Flow

Chapter Objectives To evaluate axial temperature distributions for fluids flowing inside closed conduits with either constant wall temperature or constant wall heat flux. • To generate applicable dimensionless parameters for internal forced convection using dimensional analysis. • To catalog the empirical correlations for laminar, transition, and turbulent internal flow. • To use the empirical expressions for h, along with the energy equation, to evaluate the heat transfer behavior of representative systems. •

22.1

Introduction

Heat transfer by convection occurs through the exchange of thermal energy between a surface and an adjacent fluid. Fluid motion is frequently involved in practical energy exchange processes. In this chapter, we will build upon the knowledge acquired in earlier chapters involving fluid mechanics to include the situations in which temperature differences exist between a fluid and the surface it contacts. Initially, we will treat forced convection leaving the subject of natural or free condition until Chapter 24. In this chapter, we will address the situation where the fluid, being either heated or cooled, is flowing in a closed conduit such as a pipe, tube, or duct. Fluid flow issues associated with internal flow were discussed in Chapter 16. Many of the concepts and much of the terminology introduced earlier will be used in this treatment without additional discussion. The basic rate equation for convective heat transfer, as introduced earlier, is the Newton rate equation. This relationship is ˙ = h S(TH − TC ) Q

(22.1)

˙ is the rate of heat transfer in watts, S is the surface area, TH and TC are the higher and where Q cooler temperatures, respectively, and h is the convective heat transfer coefficient in W/m2 K. Equation 22.1 is, in a practical sense, the definition of h, the heat transfer coefficient. Much of the challenge in evaluating convection heat transfer lies in the evaluation of h, which involves the complexities of fluid flow in addition to temperature distribution effects. We will examine the means for evaluating h in a later section. First, we will look at the bigger picture involving energy exchange with internal flow. 707

708

22.2

Introduction to Thermal and Fluid Engineering

Temperature Distributions with Internal Forced Convection

22.2.1 The Constant Wall Heat Flux Case Flow in a portion of a circular conduit is shown in Figure 22.1 for the condition where heat is being added to the fluid. A first law analysis of the control volume for steady-state conditions yields the expression ˙ cv + mh Q ˙ i − mh ˙ e =0

(22.2)

˙ cv is the heat transferred from the wall to the fluid Here, Q ˙ ˙ cv = Q d (d x) = q˙ d (d x) Q S ˙ where Q/S = q˙ is the heat flux, mh ˙ i is the energy entering the control volume via fluid flow mh ˙ i =

d 2 ˆ pT Vc 4

and mh ˙ e is the energy leaving the control volume by fluid flow mh ˙ e =

d 2 ˆ p (T + dT) Vc 4

Observe that, as a result of heat addition from the wall, the fluid temperature has increased by the amount, dT. With the expressions for the individual terms substituted, Equation 22.2 becomes ˙ Q d 2 d 2 ˆ pT −  ˆ p (T + dT) = 0 d(d x) +  Vc Vc S 4 4 and, after simplifying, this expression can be written as ˙ Q d dT ˆ p − Vc =0 S 4 dx

(22.3a)

˙ dT 4 Q/S − =0 ˆ p dx d Vc

(22.3b)

or

Control Volume

d V dx x FIGURE 22.1 Control volume with fluid in forced convection.

Forced Convection—Internal Flow

709

T

. dT = 4 Q/s dx d ρVcp

T(x)

Tin

x FIGURE 22.2 Axial temperature variation of a fluid being heated uniformly.

Equation 22.3b indicates that the rate of temperature increase by the fluid is proportional ˙ to the wall heat flux. If Q/S is constant, as would be true for a metal pipe wall subjected to a constant voltage source, the value of dT/d x would be a constant and the temperature variation of the fluid will be as shown in Figure 22.2. Information of interest, in addition to the fluid temperature variation, T(x), is the wall temperature. Equation 22.1 provides the means for evaluating the wall temperature, Tw (which, in this case, is TH ). The wall temperature is Tw = Tfluid +

˙ Q/S h

(22.4)

This expression indicates that we must know the value of h in order to evaluate Tw . If h is constant, Equation 22.4 indicates that the wall temperature variation will be linear with a constant slope equal to that given by Equation 22.3b. A constant value of h is generally achieved after the fluid has progressed some distance from the entrance to the conduit. Typical wall temperature variation with axial position is shown in Figure 22.3, where the distance from the entrance to the point where the two temperature profiles become parallel is designated as the thermal entrance length, L eT .

Example 22.1 Lubricating oil, to be used in an internal combustion engine test (Figure 22.4), is preheated by passing it through an electrically heated tube 1 cm in diameter and 50-cm long. The oil flows at a rate of 800 kg/h and enters at a temperature of 0◦ C. The density and specific heat of the oil are  = 836 kg/m3 and c = 2.26 J/kg-K, respectively, and the heat transfer coefficient is 68 W/m2 -K. Determine (a) the amount of energy in watts that must be transferred to the oil to achieve a temperature of 20◦ C at the tube exit and (b) the maximum tube surface temperature.

Solution Assumptions and Specifications (1) The heat transfer process is in steady state. (a) For the lubricating oil, a liquid, the specific heat is designated as c. The required heat flux can be determined from Equation 22.3a ˙ Q d dT ˆ − Vc =0 S 4 dx

(22.3a)

710

Introduction to Thermal and Fluid Engineering T Twall

Tfluid

Tin

x LeT FIGURE 22.3 Axial temperature variation of a fluid and a uniformly heated wall.

This expression can be modified by noting that ˆ = V ˆ m ˙ = AV

d 2 4

which gives   ˙ Q d 4m ˙ dT m ˙ dT = c = c 2 S 4 d dx d d x ˙ Because the heat flux, Q/S, is a constant, the temperature gradient, dT/d x, must be constant as well. Hence, we can write dT T = dx x and with this substitution, the heat flux expression is ˙ Q m ˙ T = c S d x h = 68 W/m2-K m = 800 kg/h T = 0°C ρ = 836 kg/m3 cp = 2.26 J/kg-K FIGURE 22.4 Configuration for Example 22.1.

1 cm

50 cm

Electrically Heated Tube

T = 20°C

Forced Convection—Internal Flow

711

With numerical values inserted we have ˙ Q (800 kg/h)(2.26 J/kg-K)(20 K) = = 639 W/m2 S (0.01 m)(0.50 m)(3600 s/h) The energy to the oil is thus, ˙ = (639 W/m2 )(0.01 m)(0.50 m) = 10.0 W ⇐ Q (b) Equation 22.4 will yield the maximum wall surface temperature. Because the maximum fluid temperature, at the exit, is 20◦ C, the maximum wall temperature is Tw, max = Tfluid, max + = 20◦ C +

˙ Q/S h

639 W/m2 68 W/m2 -K

= 20◦ C + 9.40◦ C = 29.4◦ C ⇐

The foregoing example dealt with the simplest case, that of a uniform heat flux. When ˙ the wall heat is known to vary along the surface, that is ( Q/S)(x) is a known function, Equations 22.3 and 22.4 still apply but their solutions will become more mathematically complex. 22.2.2 The Constant Wall Temperature Case The other boundary condition of general interest is the case where the wall temperature variation, T(x), is known. For the simplest case, where T(x) is constant, the axial temperature variations of the wall and the fluid will be as indicated in Figure 22.5. In this case, the wall temperature is higher than the entering fluid temperature, T1 . The fluid temperature will increase due to energy transport from the wall, approaching the wall temperature for large values of x. An analysis of this situation will involve the same control volume shown in Figure 22.1 with Equations 22.3 as the results. The difference between this case and the constant wall T

Twall

Tw Tfluid T1

X

FIGURE 22.5 Temperature profiles for Tw = constant and Tw > T1 .

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heat flux case is that we must now utilize Equation 22.1 for the heat flux in the form ˙ Q = h(Tw − T) S

(22.5)

The combination of Equations 22.3b and 22.5 yields dT 4 h(Tw − T) − =0 ˆ p dx d Vc

(22.6)

The solution to Equation 22.6 is made easier by defining  = Tw − T and because Tw is constant, we have dT d =− dx dx With appropriate substitution, Equation 22.6 becomes d 4 h + =0 ˆ p d x d Vc

(22.7)

For the boundary conditions (0) = 1

and

(x) = 

Equation 22.7 can be solved by separation of the variables and integration:    x d 4 h + dx = 0 ˆ p 0 d Vc 1  ln

 4 h + x=0 ˆ p 1 d Vc

This yields the result  ˆ = e −4hx/d c p V 1

(22.8)

Tw − T ˆ = e −4hx/d c p V Tw − T1

(22.9a)

or equivalently

The fluid temperature shown in Figure 22.5 is thus seen to approach the wall temperature in a decreasing exponential fashion. At some length, x = L, where T = T2 , Equation 22.9a becomes Tw − T2 ˆ = e −(4h L/d c p V) Tw − T1 ˆ the exponent For the case of a liquid where c p may be taken as c, and with m ˙ = AV, becomes =−

4h L 4h L A 4h L A =− =− ˆ ˆ d mc ˙ d c V d Ac V

Forced Convection—Internal Flow

713

But A = d 2 /4 so that 4A = d 2 and S = Ld. Hence, =−

hS mc ˙

and for the case of sized length, L , with a constant wall temperature, Tw , Equation 22.9a becomes Tw − T2 ˙ = e −(h S/mc) Tw − T1

(22.9b)

Because the wall temperature was specified in this case, the quantity of heat transferred must be determined using a simple energy balance between the entrance and any downstream location. Thus,

or

˙ = mc Q ˙ p [T(x) − T1 ]

(22.10a)

  ˆ ˙ = mc Q ˙ p 1 1 − e −(4hx/d c p V)

(22.10b)

Example 22.2 For the same circular tube as in Example 22.1 with the lubricating oil entering at 0◦ C and flowing at a rate of 800 kg/h, the wall temperature is held constant at Tw = 20◦ C. Using the value of h given in Example 22.1 (68 W/m2 -K), determine (a) the exit temperature and (b) the total heat transfer.

Solution Assumptions and Specifications (1) The heat transfer process is in steady state. (a) The exit temperature of the oil can be determined from Equation 22.9a Tw − T ˆ = e −4hx/dc V Tw − T1

(22.9a)

Because the mass flow rate is a given quantity, we can rewrite the right-hand side by substituting ˆ = V ˆ m ˙ = AV

d 2 4

and the expression becomes Tw − T ˙ = e −hd x/c m) Tw − T1 With Tw = 20◦ C and T1 = 0◦ C, the left-hand side becomes Tw − T 20◦ C − T = ◦ Tw − T1 20 C − 0◦ C and for the right-hand side with m ˙ = 800/3600 = 0.222 kg/s, we can evaluate the exponent. −

d hx (0.01 m)(68 W/m2 -K)(0.50 m) =− = −2.129 cm ˙ (2.26 J/kg-K)(0.222 kg/s)

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We can now solve for the exit temperature 20◦ C − T = e −2.129 = 2.129 20◦ C − 0◦ C which yields T = 20(1 − 0.119) = 20(0.881) = 17.6◦ C ⇐ (b) The energy transfer to the oil can be determined from an energy balance. Application of Equation 22.10a yields ˙ = mc[T(x) Q ˙ − T1 ] = (0.222 kg/s)(2.26 J/kg-K)(17.6 K) = 8.85 W ⇐ The foregoing developments and examples have presumed a knowledge of the heat transfer coefficient, h. In the next section, we will discuss the means by which it is evaluated.

22.3

Convective Heat Transfer Coefficients

Equation 22.4 defines the heat transfer coefficient, h as the ratio of convective heat flux to the temperature difference, that is, the driving force, which produces an energy exchange between a surface and an adjacent fluid. The hydrodynamic boundary layer considered in Chapter 16, is the region, close to the solid boundary, at which velocity gradients exist. When a temperature difference exists between a fluid and a surface, there is also a thermal boundary layer where, by definition, temperature gradients exist. It is within these boundary layers that the resistances to momentum and energy exchange occur. One means of evaluating h is through an analysis of the hydrodynamic and thermal boundary layers using the fundamental conservation laws. The governing relationships that must be solved are higher-order nonlinear partial differential equations, which require considerable mathematical sophistication and/or massive numerical capability to achieve useful results. Such methods are beyond the scope of this text. The equations that we will use for evaluating h are based upon empirical (experimental) results achieved by numerous researchers over the years. They have been thoroughly corroborated and are generally accepted within the technical community. Before presenting the working equations for h, we will examine the basis for using the dimensionless parameters that are involved. As discussed in Chapter 15, dimensionless analysis is a vital tool in analyzing and presenting experimental results. Table 22.1 lists the variables that are important in evaluating h, the heat transfer coefficient. Also listed are the units of each variable as well as its dimensionless representation in terms of length, L, mass, M, time, T, temperature, , and heat, Q. Note that two new fundamental dimensions,  and Q, have been added to those utilized previously. Use of the Buckingham method introduced in Chapter 15 shows that the required number of dimensionless parameters is given, in this case, by i = n−r = 8−4 = 4 The reader may verify that the rank of the dimensional matrix is 4.

Forced Convection—Internal Flow

715

TABLE 22.1

Fundamental Variables for Internal Forced Convection Variable

Symbol

Units

Dimensions

d L   cp k ˆ V

m m kg/m3 Pa-s J/kg-K W/m-K m/s W/m2 -K

L L M/L 3 M/L T Q/M Q/T L L/T Q/T L 2 

Diameter Length Fluid density Fluid viscosity Fluid specific heat Fluid thermal conductivity Velocity Heat transfer coefficient

h

ˆ as the core variables and, continuing with the Buckingham We will choose d, k, , and V method we find that the four -groups are ˆ dL 1 = d a k b c V ˆ d 2 = d a k b c V ˆ dcp 3 = d a k b c V ˆ dh 4 = d a k b c V We see by inspection that 1 =

L d

(22.11a)

or in the preferred form, which is also clearly dimensionless, 1 =

d L

Writing 2 in a dimensionless form gives us       Q b M c L d M 2 = M0 L 0 T 0 0 Q0 = L a T L LT T L3 and equating exponents on M, L , T, , and Q yields M:

0 = c+1

L:

0 = a −b−c+d −3

T:

0 = −b − c − d

:

0 = −b

Q:

0=b

The unknowns are easily evaluated as a =

1

b =0

c = −1

d=1

and 2 becomes 2 =

ˆ d V 

which we recognize as the Reynolds number.

(22.11b)

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Introduction to Thermal and Fluid Engineering

Solving for 3 and 4 in this manner yields 3 =

c p ≡ Pr k

(The Prandtl number)

(22.11c)

4 =

hd ≡ Nu k

(The Nusselt number)

(22.11d)

and

The last two groups are encountered for the first time. They are both fundamental in forced convection heat transfer. The Prandtl number, Pr, expressed by Equation 22.11c, is seen to be a combination of fluid properties. Thus, it too can be considered to be a fluid property. Physically, it is the ratio of the diffusivities of momentum and heat, /. The Nusselt number, Nu, given by Equation 22.14, is the dimensionless dependent variable because it contains the heat transfer coefficient, h, which is the objective of our analysis. In a dimensionless form we should expect to see correlations for forced convection in the form   d Nu = f , Re, Pr (22.12) L As mentioned earlier, functional forms for the right-hand side of Equation 22.12 have been evaluated using experimental results. We should note that the Nusselt number and the Reynolds number each include a length variable. Because the important length scale for internal flow is the diameter d, or the equivalent diameter if the conduit is not circular, we have currently written Nu =

hd k

and Re =

ˆ d V 

In subsequent chapters, the significant length variable may not be so obvious. To avoid confusion, we will use subscripts or Nu and Re to reflect the appropriate dimension on which they are based. Examples are Nud =

hd , k

Nux =

hx , k

and

Nu L =

hL k

and Red =

ˆ d V , 

Rex =

ˆ x V , 

and

Re L =

ˆ L V 

For the present, it should be noted that for the region near the entrance, 0 < x < L eT , the heat transfer coefficient varies with axial location. For x > L eT , the heat transfer coefficient is constant with a value that depends on the wall heating condition. For the case of constant wall heat flux, with x > L eT Nud =

hd = 4.364 k

(22.13)

and for the case of constant wall temperature with x > L eT Nud =

hd = 3.658 k

(22.14)

Forced Convection—Internal Flow

717

TABLE 22.2

Empirical Expressions for Internal Forced Convection Flow Type Laminar Transition Turbulent

Re Range

Pr Range

L/d Range

Applicable Equation

Re < 2300 2300 < Re < 104 Re > 104

Pr > 0.50 Pr > 0.50 0.70 < Pr < 160

all all L/d > 10

22.16 22.17 22.18

where for laminar flow ˆ L eH d V ≈ 0.05Red ≈ 0.05 d  and ˆ L eT d V ≈ 0.05Red Pr ≈ 0.05 Pr d 

(22.15a)

L eH = L eT = 10d

(22.15b)

For turbulent flow

Table 22.2 lists the empirical expressions that can be used for evaluating Nu, or h, for various internal flow cases. The equations to be used for each case are provided in Sections 22.3.1 through 22.3.3. 22.3.1 Case 1—Laminar Flow The widely used Sieder-Tate equation (1936) is Nud =

1/3 1.86Red Pr1/3

 1/3  0.14 d b L w

(22.16)

The ratio, b /w , involves the viscosity at the temperature of the bulk fluid, b , and the viscosity at the wall temperature, w . All other fluid properties are evaluated at the bulk fluid temperature. The reader may recall that, for internal flow, the hydrodynamic entrance length is defined as the distance between the entrance to the downstream location where the flow becomes fully developed. As noted earlier in the present chapter, a thermal boundary layer develops in internal flow when the fluid is heated or cooled by contact with the wall of the conduit. One should, thus, expect there to be a thermal entrance length analogous to the hydrodynamic case, and this is indeed true. The concept of a thermal entrance length is, however, not so obvious; a fluid will continue to be heated or cooled as it progresses through a conduit—even after the thermal entrance length has been exceeded. Analytical solutions have been achieved for laminar flow cases. A classic solution exists for internal flow with the velocity profile fully developed (parabolic) when the wall temperature is suddenly raised and maintained constant for x > L eT . This is the Graetz problem first solved in 1885. Other variations include cases of both the velocity and temperature profiles developing simultaneously, with the wall condition being one of specified heat flux rather than temperature. The interested reader may consult Welty et al. (2008) or Kays and Crawford (1980) for a more detailed discussion.

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As a practical matter, the Sieder-Tate equation, which is empirical, does a satisfactory job of predicting h for laminar internal flow, and its use is widely accepted within the heat transfer community. 22.3.2 Case 2—Transition Flow The correlation of Hausen (1943) is recommended.  2/3  Nud = 0.116 Red − 125 Pr1/3



b w

0.14

 2/3

d 1+ L

(22.17)

Fluid properties are evaluated at the bulk fluid temperature in all cases except w , which is evaluated at the wall temperature. 22.3.3 Case 3—Turbulent Flow The Dittus-Boelter (1930) equation is n Nud = 0.023Re0.8 d Pr

(22.18)

where the exponent, n, is taken as n = 0.40 (Tw > Tb ) n = 0.30 (Tw < Tb ) Fluid properties are evaluated at the bulk fluid temperature.

22.4

Applications of Internal Flow Forced Convection Correlations

Examples 22.3 through 22.5 illustrate the use of the capabilities developed in the preceding sections.

Example 22.3 Oil at 300◦ C is heated by steam condensing at 370◦ C on the outside of a steel pipe with an outside diameter (OD) of 2.67 cm and an inside diameter (ID) of 2.09 cm. The pipe is 2.5-m long and the mass flow rate through the pipe is 0.245 kg/s. Properties of the oil are as tabulated: T, K

ρ, kg/m3

cp , J/kg-K

k, W/m-K

μ Pa-s

300 310 340 370

910 897 870 865

1.84 × 103 1.92 × 103 2.00 × 103 2.13 × 103

0.133 0.131 0.130 0.128

41.4 × 103 22.8 × 103 7.89 × 103 3.72 × 103

Determine (a) the temperature of the oil as it leaves the pipe and (b) the heat transferred.

Forced Convection—Internal Flow

719 Steam Condensing at 370°C Pipe OD = 2.67 cm ID = 2.09 cm L = 2.5 m

At 370°C, μ = 0.00372 kg/m-s

0.245 kg/s Oil at 300°C

FIGURE 22.6 Configuration for Example 22.3.

Solution Assumptions and Specifications (1) Negligible temperature drop occurs across the tube wall so that we are dealing with a constant inside surface temperature of 370◦ C. (2) The heat transfer process is in the steady state. In this constant surface temperature situation Equation 22.9a applies and, because the fluid properties are temperature dependent and the exit temperature of the oil is unknown, a trial-and-error approach will be required. We will assume an exit temperature for the initial evaluation of fluid properties. This assumption will be checked and the solution repeated until sufficient agreement is achieved. (a) The pipe is sketched in Figure 22.6. We will employ Equation 22.9a Tw − T ˆ = e −4hx/d c p V Tw − Tin

(22.9a)

We observe that the evaluation of the heat transfer coefficient, h, requires a knowledge of the flow regime which, in turn requires a knowledge of the Reynolds number. Since the mass flow rate, m, ˙ is specified, the expression for Re is Red =

ˆ dV md ˙ 4md ˙ 4m ˙ = = =  A d 2 d

Initially, an exiting fluid temperature of 310 K will be assumed and the corresponding average or bulk temperature of the oil is 305◦ C. At this temperature, the properties of the oil are c = 1880 J/kg-K,

 = 0.0321 Pa-s,

and

k = 0.132 W/m-K

and assuming that the pipe wall is at the temperature of the condensing steam, we have w = 0.00372 Pa-s

720

Introduction to Thermal and Fluid Engineering

The Reynolds number is ˆ d V 4m ˙ 4(0.245 kg/s) = = = 465  d (0.0209 m)(0.0321 Pa-s) This result indicates that the flow is laminar, which dictates that Equation 22.16 applies and we can write hd 1/3 = 1.86Red Pr1/3 k

 1/3  0.14 d b L w

(22.16)

The Reynolds number has already been determined. The Prandtl number is evaluated as Pr =

c (0.0321 kg/m-s )(1880 J/kg-K) = = 457 k 0.132 W/m-K

Substituting numerical values into Equation 22.16 yields  h=

 0.132 W/m-K (1.86)(465) 1/3 (457) 1/3 0.0209 m 

0.0209 m × 2.5 m

1/3 

0.0321 kg/m-s 0.00372 kg/m-s

0.14

or h = 192.4 W/m2 -K ˆ to With this result, we can solve Equation 22.9a, which may be modified with m ˙ = AV the form Tw − T ˙ = e −h Ld/mc Tw − T1 Inserting appropriate numerical values we obtain h Ld (192.4 W/m2 -K)(2.5 m)(0.0209 m) = = 0.0686 mc ˙ (0.245 kg/s)(1880 J/kg-K) and, finally, T − 370◦ C = e −0.0686 = 0.934 300◦ C − 370◦ C The temperature at the exit of the pipe is thus, T = 370◦ C − (0.934)(70◦ C) = 370◦ C − 65.4◦ C = 304.6◦ C This result is lower than the assumed value of 310◦ C. At an adjusted value of the exit temperature of 304◦ C, we have a bulk temperature 302◦ C and adjusted properties of c = 1858 J/kg-K,

 = 0.0377 kg/m-s,

and k = 0.133 W/m-K

Forced Convection—Internal Flow

721

The updated results are ˆ d V 4m ˙ 4(0.245 kg/s) = = = 396  d 0.0209 m)(0.0377 kg/m-s) Pr =

c (0.0377 kg/m-s)(1858 J/kg-K) = = 527 k 0.133 W/m-K 

h=

 0.133 W/m-K (1.86)(396) 1/3 (527) 1/3 0.0209 m 

×

0.0209 m 2.5 m

1/3 

0.0377 kg/m-s 0.00372 kg/m-s

0.14

or h = 197.1 W/m2 -K Then h Ld (197.1 W/m2 -K)(2.5 m)(0.0209 m) = = 0.0715 mc ˙ (0.245 kg/s)(1858 J/kg-K) T − 370◦ C = e −0.0715 = 0.931 300◦ C − 370◦ C and, the temperature at the exit of the pipe is T = 370◦ C − (0.931)(70◦ C) = 370◦ C − 65.2◦ C = 304.8◦ C ⇐ This is sufficiently close to the initial assumption. Hence, another calculation is unnecessary. (b) The heat received by the oil will be ˙ = mc(T Q ˙ exit − T1 ) = (0.245 kg/s)(1858 J/kg-K)(304.8◦ C − 300◦ C) = 2182 W ⇐

Example 22.4 Hot water flows through a 1.6-cm-internal diameter pipe (Figure 22.7) at a

velocity of 0.10 m/s. Its temperature at the pipe inlet is 80◦ C and the pipe wall temperature remains constant at 30◦ C. Determine (a) the temperature of the water at the exit from the pipe and (b) the quantity of heat lost in a waterline that is 1.55-m long.

Solution Assumptions and Specifications (1) The wall temperature is specified constant at 30◦ C. (2) The heat transfer process is in the steady state.

722

Introduction to Thermal and Fluid Engineering Tw = 30°C

Water 80°C V = 0.10 m/s

1.6 cm

1.55 m FIGURE 22.7 Configuration for Example 22.4.

Equation 22.9a applies in this case Tw − T ˆ = e −4hx/d c V Tw − T1

(22.9a)

A value for the heat transfer coefficient, h, is needed. The first task is to evaluate the Reynolds number to determine which equation for h applies. To get a handle on Re, fluid properties at Tb = 60◦ C, which is consistent with an exit temperature of 40◦ C, will be used. Table A.14 in Appendix A shows that at 60◦ C (333 K),  for water = 0.483 × 10−6 m2 /s. The Reynolds number is thus determined as Red =

ˆ dV (0.016 m)(0.10 m/s) = = 3313  0.483 × 10−6 m2 /s

The water flow appears to be in the transition region that is Case 2 in Table 22.1. Equation 22.17 applies  0.14  2/3

b d 2/3 1/3 Nud = 0.116[Red − 125]Pr 1+ (22.17) w L For a first set of calculations, we will use Tb = 60◦ C. The properties of water obtained from Table A.14 are  = 983.3 kg/m3 c = 4183 J/kg-K  = 475 × 10−6 Pa-s

k = 0.654 W/m-K Pr = 3.04 w = 803 × 10−6 kg/m-s

and after appropriate substitution, with k 0.654 W/m-K = = 40.88 W/m2 -K d 0.016 m  2/3   d 0.016 m 2/3 1+ =1+ = 1 + 0.0474 = 1.0474 L 1.55 m and 

b w

0.14

 =

475 × 10−6 kg/m-s 803 × 10−6 kg/m-s

0.14 = (0.592) 0.14 = 0.929

Forced Convection—Internal Flow

723

we obtain h = (40.88)(0.116)[(3313) 2/3 − 125](3.04) 1/3 (0.929)(1.0474) = 650 W/m2 -K The argument of the exponential term in Equation 22.9a is evaluated as 4 hL 4 (650 W/m2 -K)(1.55 m) = = 0.612 ˆ d c p V 0.016 m (983.3 kg/m3 )(4183 J/kg-K)(0.10 kg/s) which yields T2 − 30◦ C = e −0.612 = 0.542 80◦ C − 30◦ C and T2 = 30◦ C + (0.542)(50◦ C) = 30◦ C + 27.1◦ C = 57.1◦ C This value differs significantly from the assumed value of 40◦ C and we must redo the calculations. With the exit temperature next assumed to be 56◦ C with Tb = 68◦ C, the fluid properties obtained from Table A.14 (with interpolation) are  = 978.9 kg/m3 c p = 4186 J/kg-K  = 421 × 10

−6

Pa-s

k = 0.662 W/m-K Pr = 2.67 w = 803 × 10−6 kg/m-s

The Reynolds number will be Red =

ˆ dV (0.016 m)(978.9 kg/m3 )(0.10 m/s) = = 3720  421 × 10−6 Pa-s

After appropriate substitution, with k 0.662 W/m-K = = 41.38 W/m2 -K d 0.016 m and



b w

0.14

 =

421 × 10−6 kg/m-s 825 × 10−6 kg/m-s

0.14 = (0.510) 0.14 = 0.910

we obtain h = (41.38)(0.116)[(3720) 2/3 − 125](2.67) 1/3 (0.910)(1.0474) = 730.4 W/m2 -K Then, with the exponential argument determined as 4 h 4 (730.4 W/m2 -K)(1.55 m) L= = 0.691 ˆ d c p V 0.016 m (978.4 kg/m3 )(4186 J/kg-K)(0.10 kg/s) we have T2 − 30◦ C = e −0.691 = 0.501 80◦ C − 30◦ C

724

Introduction to Thermal and Fluid Engineering

and T2 = 30◦ C + (0.501)(50◦ C) = 30◦ C + 25◦ C = 55◦ C ⇐ This value is sufficiently close to our last assumed value that we may consider it to be our desired result. (b) With m ˙ = AV, the heat lost by the water will be ˙ = AVc ˆ p (T2 − T1 ) Q and with T2 − T1 = 80◦ C − 55◦ C = 25◦ C  (0.016 m) 2 3 ˙ Q = (978.9 kg/m ) (0.10 m/s)(4186 J/kg-K)(25◦ C) 4 = 2060 W ⇐

Example 22.5 Air at atmospheric pressure and 10◦ C enters a 10-m-long rectangular duct

(Figure 22.8) whose walls are maintained at 70◦ C by solar irradiation. The duct is 15-cm wide by 7.5-cm high. Determine the flow rate of the air if its exit temperature is to be 50◦ C.

Solution Assumptions and Specifications (1) The wall temperature is specified constant at 70◦ C. (2) The air inlet and outlet temperatures are specified as 10◦ C and 50◦ C, respectively. (3) The duct is specified as rectangular. (4) The heat transfer process is in the steady state. Initially, we will assume turbulent flow and this assumption will be validated as we proceed. The situation, as shown, is one of internal flow with a constant wall temperature. Equation 22.9a Tw − T ˆ = e −4hx/d c p V Tw − T1

(22.9a)

50°C 70°C

10 m 7.5 cm Air Atmospheric Pressure 10°C FIGURE 22.8 Configuration for Example 22.5.

15 cm

Forced Convection—Internal Flow

725

is therefore applicable and we can evaluate the left-hand side by inserting the specified temperature values Tw − T2 70◦ C − 50◦ C 20◦ C = ◦ = ◦ = 0.333 ◦ Tw − T1 70 C − 10 C 60 C With this value, Equation 22.9a yields e −4hx/d c p V = 0.333 ˆ

which can be simplified to −

4 hx = ln 0.333 = −1.099 ˆ d c p V

or h = 1.099

ˆ dc p V 4x

(a)

We note that a diameter is involved in this expression even though the duct has been specified as having a rectangular cross section. This is handled by employing the equivalent diameter that was introduced in Chapter 15. For this problem, the equivalent diameter is de =

4A 4(0.15 m)(0.075 m) = = 0.10 m P 2(0.15 m + 0.075 m)

Because turbulent flow was assumed, we choose Equation 22.18 (the Dittus-Boelter equation) as the correlating equation. For Tw > Tb , n = 0.40

0.8 ˆ hd de V 0.8 0.4 Nud = Pr0.4 = 0.023Red Pr = 0.023 k  The properties, k, , , and Pr are temperature dependent. At the average bulk temperature of 30◦ C (303 K), the property values, obtained from Table A.13 (with interpolation), are  = 1.167 kg/m3

k = 0.0264 W/m-K c p = 1005.9 J/kg-K

Pr = 0.707

 = 1.86 × 10−5 Pa-s With appropriate substitution we obtain k h = 0.023 de

ˆ de V 

0.8



Pr0.4

0.0264 W/m-K = 0.023 0.10 m ˆ 0.8 W/m2 -K = 5.77V



ˆ (0.10 m) V(1.167 kg/m3 ) 1.86 × 10−5 kg/m-s

0.8 (0.707) 0.4

726

Introduction to Thermal and Fluid Engineering

and from Equation (a), we have h = 1.099 = 1.099

ˆ de c p V 4x ˆ (0.10 m)(1.167 kg/m3 )(1005.9 J/kg-K) V 4(10 m)

ˆ W/m2 -K = 3.225V The two expressions for h may be combined ˆ 0.80 = 3.225V ˆ 5.77V ˆ 0.20 = 1.790 V ˆ = 18.38 m/s V and a check of the Reynolds number provides Re =

ˆ e Vd (1.167 kg/m3 )(18.38 m/s)(0.10 m) = = 115,320  1.86 × 10−5 kg/m-s

This indicates that the flow is well into the turbulent range. Hence, our preliminary assumption of turbulent flow was correct and the use of Equation 22.18 was appropriate. The problem may now be completed by determining m ˙ ˆ = (1.167 kg/m3 )(0.15 m)(0.075 m)(18.38 kg/s) = 0.241 kg/s ⇐ m ˙ = AV

22.5

Design Example 13

A copper tube of length, L, is to be welded to the underside of a flat plate (Figure 22.9) solar collector. The plate absorbs heat such that a uniform temperature of 85◦ C is maintained. The tube has an inner diameter of 1.25 cm and is to be designed so that it will heat 1.262 ×10−5 m3 /s of water from 15◦ C to 75◦ C. Assuming laminar flow in the tube, determine the necessary length of the tube to accomplish the desired temperature rise of the water by comparing the results using two approaches: (a) assuming the fully developed value for the Nusselt number over the entire tube length and (b) using the Sieder-Tate correlation (Equation 22.16) to determine the Nusselt number.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is no temperature drop across the tube wall. (3) The effect of bends in the tube is negligible. The first step is to confirm that the flow in the tube is indeed laminar. The volumetric flow rate is ˙ = 1.262 × 10−5 m3 /s V

Forced Convection—Internal Flow

727

Water Out

Solar Collector Panel

Water In

FIGURE 22.9 Configuration for Design Example 13.

and with A=

 2  d = (0.0125 m) 2 = 1.2272 × 10−4 m2 4 4

we have ˆ = V

˙ V 1.262 × 10−5 m3 /s = = 0.1028 m/s A 1.2272 × 10−4 m2

With Tb =

T1 + T2 15◦ C + 75◦ C = = 45◦ C 2 2

Table A.14 gives (with interpolation)  = 990.3 kg/m3

Pr = 3.905

 = 0.596 × 10−3 kg/m-s

k = 0.6385 W/m-K

c = 4.177 kJ/kg-K Thus, the Reynolds number is ˆ ˆ d V Vd (990.3 kg/m3 )(0.1028 m/s)(0.0125 m) = = = 2135   0.596 × 10−3 kg/m-s This shows that the flow in the tubing is indeed laminar. Moreover, ˆ = (990.3 kg/m3 )(1.2272 × 10−4 m2 )(0.1028 m/s) = 0.0125 kg/s m ˙ = AV Then, e − =

Tw − T2 85◦ C − 75◦ C = ◦ = 0.1429 Tw − T1 85 C − 15◦ C

so that − =

hS = ln 0.1429 = −1.9456 mc ˙

728

Introduction to Thermal and Fluid Engineering

or  = 1.9456 This makes the h S product h S = (1.9456) mc ˙ = (1.9456)(0.0125 kg/s)(4177 J/kg-K) = 101.6 W/K This h S product forms the underpinning for both parts (a) and (b) which now follow. (a) Equation 22.14 gives the fully developed value for h/d k hd = 3.658 k Then h=

k hd = d k



 0.6385 W/m-K 3.658 = 186.9 W/m2 -K 0.0125 m

and to achieve an h S product of 101.6 W/K, we have h S = 101.6 W/K = hd L = (186.9 W/m2 -K)(0.0125 m)L or L=

101.6 W/K (181.6 W/m2 -K)(0.0125 m)

= 14.25 m ⇐

(b) The heat transfer coefficient via the Sieder-Tate correlation may be obtained from Equation 22.16.     1/3  0.14 ˆ k hd b k d V d h= = 1.86 Pr d k d  L w and with S = d L, the h S product will be

    1/3  0.14 ˆ k d V d b h S = 101.6 W/K = 1.86 Pr dL d  L w

= 1.86k

ˆ d V Pr d 

1/3 

b w

0.14 L 2/3

The length required to yield h S = 101.6 W/K will be

3/2 101.6 W/K L= ˆ 1.86k(Vd/)Pr d 1/3 (/w ) 0.14 At Tw = 85◦ C, w = 0.0.3385 × 10−3 kg/m-s. Hence, with ˆ d V Pr d = (2135)(3.905)(0.0125 m) = 104.21 m  and b 0.596 × 10−3 kg/m-s = = 1.7607 w 0.3385 × 10−3 kg/m-s

Forced Convection—Internal Flow

729

we obtain 

101.6 L= 1.86(0.6385 W/m-K)(104.21 m) 1/3 (1.7607) 0.14

3/2 = 12.36 m ⇐

The results of parts (a) and (b) indicate that entrance lengths are significant for this case. Use of the fully developed value for Nu gives a result for L, which is approximately 10% larger than the value obtained using the Sieder-Tate correlation. With an eye toward being conservative, we will choose the length of tubing to be L = 14.25 m ⇐ and we observe that, for fluids with appreciably larger values of Pr, the error in using the fully developed limit can be quite large (in excess of 200% in some cases). We turn now to a design example involving a high-heat dissipating electronic component. The traveling wave tube is an electron tube in which a stream of electrons interacts continuously (or repeatedly) with a guided electromagnetic wave moving substantially with it. The interaction occurs in such a way that there is a net transfer of energy from the stream of electrons to the guided electromagnetic wave. The tube is used as an amplifier or oscillator at frequencies in the microwave region. An artist’s conception of the anode (collector) end of a traveling wave tube is shown in Figure 22.10. Notice that the entire configuration is very small and, because the heat dissipation is substantial, the heat release in a traveling wave tube collector can approach that of a steam generator in a modern steam power plant. The electrons are “collected” in the small cavity at the center of the configuration and, because of the high-heat release, the collector must be liquid cooled. Several liquids are available and they all meet the requirements of high-dielectric strength, chemical inertness, thermal stability, effects of moisture, pour and flash points, flammability, toxicity, surface tension, and vapor pressure. Water is not suitable as a liquid coolant because of its low-dielectric strength.

Collector

Collector Cavity

Tube Structure

FIGURE 22.10 Artist’s conception of the collector end of a traveling wave tube.

730

Introduction to Thermal and Fluid Engineering TABLE 5.1

Performance of a Liquid-Cooled System Dissipating 1500 W Assume m ˙ T1 Assume T2 Tb c at Tb ˙ mc T = Q/ ˙ T2 T2 OK? Tb c at Tb ˙ mc T = Q/ ˙ T2 T2 OK?

kg/s ◦C ◦C ◦C J/kg-K ◦C ◦C ◦C

J/kg-K ◦C ◦C

0.100 77.0 84.6 80.8 2093 7.2 84.2 no 80.6 2092 7.2 84.2 Yes

0.125 77.0 83.0 80.0 2090 5.7 82.7 no 79.9 2089 5.7 82.7 Yes

0.150 77.0 81.8 79.4 2088 4.8 81.8 Yes

0.175 77.0 81.0 79.0 2087 4.1 81.1 Yes

0.200 77.0 80.6 78.8 2086 3.6 80.6 Yes

The determination of the anode (collector) temperature using a liquid coolant requires an input from at least two of the disciplines treated in this book: 1. It begins with a first law consideration (Chapter 5) in which the range of liquid flows to accommodate the specified heat dissipation are established. This was performed as Design Example 1 and the results were summarized in Table 5.1, which is repeated here for the reader’s convenience. 2. The surface temperature of the collector wall is then obtained using procedures found here in Chapter 22.

22.6

Design Example 14

A traveling wave tube collector is to be cooled in an application where the uniform dissipation in the collector is 1.5 kW (1500 W). Coolanol-45 (Monsanto Chemical Co.), which enters the collector at 77◦ C, is to be used. The collector material is copper (k = 385 W/m-K) and in order to prevent loss of vacuum due to “outgassing” of the copper material, the walls of the collector must be held to a maximum temperature of 150◦ C. The details of the collector configuration are indicated in Figure 22.11 and the properties of Coolanol-45 are provided in Table 22.4. Because it is deemed risky to interpolate the dynamic viscosity data, a plot of dynamic viscosity as a function of temperature is provided in Figure 22.12. Notice that 16 fins will be employed and the 1500 W may be considered as flowing radially outward from the collector cavity, with a diameter of 0.625 cm, to the coolant passage. There are four fins per pass in this four-pass collector and relevant dimensions include: di = 2.54 cm

do = 5.715 cm

fin height, b = 1.875 cm

fin thickness, = 2.381 mm

L = 8.255 cm

thermal conductivity, k = 385 W/m-K

number of fins = 16

fins per pass = 4

We will need to determine several pertinent parameters. These are the passage flow area, the passage wetted perimeter, the passage equivalent diameter, the finned surface area, and the base surface area.

Forced Convection—Internal Flow

Tube Body

T u r n S p a c e

731

Fin Space

Collector Fin Space

T u r n

Coolant in

S p a c e

Coolant out

8.255 cm Coolant in

2.54 cm 5.715 cm

2.381 mm

Coolant out

16 Fins 2.381-mm Thick–1.5875-cm High per Pass Fins are Copper 4 Passes FIGURE 22.11 Details of the collector of a traveling wave tube.

Strategy 1. A modification of Equation 20.21 will yield the maximum temperature allowed at the inner surface of the coolant passage. 2. The procedure is then straightforward •

For each flow rate, we determine the pertinent fluid properties, the Reynolds number, and the heat transfer coefficient. Because the flow in each case is laminar, Equation 22.16 is employed.

•

After the heat transfer coefficient, h, is obtained, the fin performance parameter, m, is determined.  m=

2h k

1/2

Then, with this value of m, the fin efficiency is obtained from Equation 20.67

=

tanh mb mb

732

Introduction to Thermal and Fluid Engineering

TABLE 22.3

Summary of Calculations for the Surface Temperature of the Finned Passage in Figure 22.11: In All Cases, T1 = 77◦ C Assume m ˙ T2 Tb c at Tb  × 10−3 at Tb Pr at Tb  at Tb  × 10−3 at 140◦ C ˆ V ˆ e / Red = Vd /w (/w ) 0.14 de /L Red Prde /L (Red Prde /L) 1/3 Nud h m2 m mb tanh mb

f Sb

f Sf S = Sb + f S f  = h S/mc ˙ F = e F T2 F T2 − T1 F −1 Tw

kg/s

0.100

0.125

0.150

0.175

0.200

◦C

84.2 80.6 2092 4.106 63.2 858.1 2.0126 0.3207 563.8 2.0267 1.1040 0.1019 3631.2 15.37 31.56 510.2 1113.2 33.36 0.5297 0.4841 0.916 0.00344 0.03840 0.04184 0.1020 1.1074 93.2 16.2 0.1074 151.2

82.7 79.9 2089 4.144 63.6 858.7 2.0126 0.4005 698.7 2.0454 1.1054 0.1019 4528.1 16.54 34.02 550.0 1200.0 34.64 0.5499 0.5005 0.910 0.00344 0.03816 0.04160 0.0876 1.0916 90.3 13.3 0.0916 145.0

81.8 79.4 2088 4.196 64.4 858.9 2.1026 0.4805 827.4 2.0711 1.1073 0.1019 5430.0 17.58 36.21 585.4 1277.2 35.74 0.5673 0.5134 0.905 0.00344 0.03795 0.04139 0.0774 1.0805 88.4 11.4 0.0805 141.4

81.1 79.0 2087 4.237 65.0 859.2 2.1026 0.5604 956.0 2.0913 1.1088 0.1019 6332.2 18.50 38.16 616.9 1346.0 36.69 0.5824 0.5244 0.900 0.00344 0.03775 0.04129 0.0697 1.0722 87.0 10.0 0.0722 137.9

80.6 78.8 2086 4.258 65.3 859.3 2.1026 0.6404 1087.2 2.1017 1.1096 0.1019 7234.6 19.34 39.92 645.4 1408.1 37.53 0.5957 0.5340 0.896 0.00344 0.03759 0.04103 0.0635 1.0656 85.9 8.9 0.0656 135.5

◦C

J/kg-K kg/m-s kg/m3 kg/m-s m/s

W/m2 -K m−2 m−1

m2 m2 m2 ◦C ◦C ◦C

TABLE 22.4

Properties of Coolanol-45: Data Extracted from Monsanto Bulletin AV-3 and Converted to SI Units ◦C

k W/m-K

μ × 10−3 kg/m-s

ν × 10−5 m2 /s

ρ kg/m3

c kJ/kg-K

Pr

40 50 60 70 80 90 100 110 120 130 140

0.132 0.133 0.135 0.136 0.136 o.135 0.133 0.132 0.130 0.127 0.125

9.721 8.185 6.201 5.167 4.134 3.669 3.101 2.687 2.356 2.191 2.026

1.086 0.932 0.712 0.597 0.482 0.431 0.367 0.320 0.283 0.265 0.247

894.9 877.8 871.4 865.8 858.5 852.1 845.7 839.3 832.8 826.4 820.1

1.954 1.988 2.022 2.056 2.090 2.124 2.158 2.192 2.226 2.260 2.294

143.9 122.3 92.9 78.1 63.5 57.7 50.3 44.6 40.3 39.0 37.2

T

Forced Convection—Internal Flow

733

Dynamic Visosity×103 kg/ms

10

8

6

4

2

40

60

80

100

120

140

Temperature, °C FIGURE 22.12 Dynamic viscosity of Coolanol-45 as a function of temperature.

•

The total surface will be S = Sb + S f and then, in each case, using Equation 22.9b, we form the parameter, F , defined as ˙ F = e −h S/mc

The temperature at the inner surface of the annular region is then found from a modification of Equation 22.9b Tw =

F T2 − T1 F−1

3. The last step is the subtraction of the temperature drop through the copper collector from the collector cavity to the inside surface of the coolant passage. 4. The results for the five flow rates may be plotted to yield the solution. Now that the strategy has been outlined, the first step is the determination of the temperature allowed at the inside surface of the coolant passage. The maximum temperature permitted in the collector is 150◦ C and it will exist at the walls of the collector cavity. A modification of Equation (20.30) will provide the maximum temperature allowed at the inner surface of the coolant passage. With dc and Tc used to respectively designate the diameter

734

Introduction to Thermal and Fluid Engineering

and temperature of the cavity Tc − To =

˙ ln do /dc Q 2k L

To = Tc −

˙ ln do /dc Q 2k L

= 150◦ C −

(1500 W)(ln 2.54 cm/0.625 cm) 2(385 W/m-K)(0.08255 m)

= 150◦ C − 10.4◦ C = 139.6◦ C Supporting data for Table 22.3 include: Flow Area   1  2 A= do − di2 − n f b 4 4   1   = (5.715 cm) 2 − (2.54 cm) 2 − (16 fins)(1.5875 cm)(0.2381 cm) 4 4 =

1 (20.585 cm2 − 6.048 cm2 ) = 3.6343 cm2 4

or A = 3.6343 × 10−4 m2 Wetted Perimeter P= =

 1 (do + di ) − 2n f + 2n f b 4 1 [(5.715 cm + 2.540 cm) − 2(16 fins)(0.2381 cm) 4 +2(16 fins)(1.5875 cm)]

=

1 (25.9338 cm − 7.6192 cm + 50.800 cm) = 17.2787 cm 4

or P = 0.1728 m The Passage Equivalent Diameter de =

4A 4(3.6343 × 10−4 m2 ) = = 8.4127 × 10−3 m P 0.1728 m

The Surface Area of the Sixteen Fins S f = 2Ln f b = 2(8.255 cm)(16 fins)(1.5875 cm) = 419.34 cm2 or S f = 0.04193 m2

Collector Temperature, °C

Forced Convection—Internal Flow

735

160

150

140

139.6°C

0.161 kg/s

130

0.10

0.12

0.14

0.16

0.18

0.20

Mass Flow Rate, m, kg/s FIGURE 22.13 Collector temperature as a function of Coolanol-45 flow rate.

The Surface Area of the Inner Portion of the Coolant Passage Sb = (di − n f )L = [(2.54 cm) − (16 fins)(0.2381 cm)](8.255 cm) = (7.9796 cm − 3.8096 cm))(8.255 cm) = 34.4237 cm2 The actual calaculations are displayed in Table 22.3 and a plot of the collector surface temperature as a function of the Coolanol-45 mass flow rate is shown in Figure 22.13. From this plot, we are able to deduce that the mass flow rate of the Coolanol-45 will be approximately 0.161 kg/s.

22.7

Summary

This chapter has dealt with convective heat transfer when the fluid flow is confined within closed conduits such as tubes, pipes, or ducts. A first law of thermodynamics application to internal flow provided the characteristic behavior of the fluid in terms of its axial temperature variation and total heat exchange for wall conditions of both constant wall temperature and constant heat input. Dimensional analysis generated the parameters that apply to internal forced convection. These parameters are •

ˆ The Reynolds number, Re = d V/

The Prandtl number, Pr = c p /k • The Nusselt number, Nu = hd/k •

736

Introduction to Thermal and Fluid Engineering

Empirical equations were introduced and their applications were demonstrated. The expressions that apply are •

For laminar flow (Re < 2300) hd 1/3 = 1.86Red Pr1/3 k

•

•

 1/3  0.14 d  L w

(22.16)

For transition flow (2300 < Re < 104 )

 2/3  0.14 hd d  2/3 1/3 = 0.116[Red − 125]Pr 1+ k L w

(22.17)

For turbulent flow (Re > 104 ) hd n = 0.023Re0.80 d Pr k

(22.18b)

where n = 0.4 for Tw > Tb and n = 0.3 for Tw < Tb .

22.8

Problems

Flow in Circular Tubes and Pipes 22.1: Lubricating oil (approximately SEA 50) flows in a 4-m-long pipe having an inner diameter of 2.50 cm at a velocity of 0.42 m/s. The oil has a mean bulk temperature of 100◦ C and the wall is at 40◦ C. Determine the heat transfer coefficient. 22.2: Air at 2.5 atm and a velocity of 6 m/s flows in 1.25-m pipe having an inner diameter of 2.50 cm. The air has a bulk temperature of 250◦ C and the wall is at 80◦ C. Determine the heat transfer coefficient. 22.3: Nitrogen at 1 atm flows in a 2-in-schedule 80 pipe at a flow rate of 12 m/s. The bulk temperature of the nitrogen is 27◦ C and the wall temperature is 77◦ C. Determine the heat transfer coefficient. 22.4: Rework Problem 22.3 if the fluid is carbon dioxide. 22.5: Rework Problem 22.3 if the fluid is air. 22.6: Water at a bulk temperature of 27◦ C enters a 1 in × 14 Birmingham wire gage tube at a velocity of 3.25 m/s. The tube is 1 m in length and the wall temperature is 47◦ C. Determine the heat transfer coefficient. 22.7: Rework Problem 22.6 for glycerin. 22.8: Rework Problem 22.6 for ethylene glycol. 22.9: Water at a velocity of 0.04 m/s flows through a 2-m-long tube having an inner diameter of 3 cm. The bulk temperature of the water is 27◦ C and the tube wall is at 80◦ C. Determine the heat transfer coefficient. 22.10: Air at 1 atm flows through a circular conduit that is 6.25 cm in inner diameter and 1.25-m long. The air has a bulk temperature of 27◦ C and the wall of the conduit is at 77◦ C. The mass rate of flow is m ˙ = 5.25 kg/h. Determine the heat transfer coefficient. 22.11: Superheated steam at atmospheric pressure and a bulk temperature of 477◦ C enters a 2-m long 8-in-schedule 40 pipe at 2.375 m/s. The pipe wall temperature is 427◦ C. Determine the heat transfer coefficient.

Forced Convection—Internal Flow

737

22.12: Exhaust gases (which may be assumed to have the properties of air) enter the tubes of an air heater at 527◦ C and discharge to the atmosphere at 27◦ C. The air heater, which is vertical and 7.5-m high, contains tubes that have inner diameters of 7.5 cm. The exhaust gas flow rate is m ˙ = 0.0625 kg/s. Determine the heat transfer coefficient. 22.13: Oxygen at 700 kPa and a flow rate of 3.50 m/s enters a pipe with an inner diameter of 1.905 cm at 17◦ C and leaves at 37◦ C. The pipe wall is at 77◦ C. Determine the heat transfer coefficient. 22.14: Ethylene glycol is heated from 17◦ C to 37◦ C in a tube whose inner diameter is 3.5 cm. The tube wall is held at 77◦ C and the velocity in the tube is 3.125 m/s. Estimate the length of the tube to ensure that h, the heat transfer coefficient, is 1325 W/m2 -K. 22.15: Air at 1 atm flows through a circular conduit at a volumetric flow rate of 1.125 m3 /s; the inner diameter of the conduit is 20 cm. The bulk temperature of the air is 15◦ C and the wall temperature is 77◦ C. Determine the heat transfer coefficient. 22.16: Lubricating oil (approximately SAE 50) with a bulk temperature of 160◦ C flows at a velocity of 0.32 m/s through a pipe with an inner diameter of 2.5 cm. The pipe wall temperature is 140◦ C and the tube length is 1 m. Determine the heat transfer coefficient. 22.17: Glycerin with a bulk temperature of 47◦ C flows at a velocity of 0.50 m/s through a pipe with an inner diameter of 5 cm. The pipe wall temperature is 27◦ C. Determine the heat transfer coefficient. 22.18: In a chilled water air-conditioning system, water enters and leaves a pipe having an inner diameter of 3.5 cm at 15◦ C and 5◦ C, respectively. The velocity of the water is 0.75 m/s, and the pipe wall temperature is 2◦ C. Determine the heat transfer coefficient. Heat Transfer Coefficients: Constant Wall Temperature and Uniform Heat Flux 22.19: Water enters a 2-in-schedule 40 pipe at 80◦ C and a flow rate of 50,400 kg/h. The pipe wall temperature is constant at 140◦ C. Determine the length of pipe necessary to allow the water to leave at 120◦ C. 22.20: For the conditions of Problem 22.19, assume that the pipe wall is subjected to a uniform heat flux. Determine the length of pipe required to hold the maximum pipe wall temperature to 150◦ C. 22.21: Water at 57◦ C and a flow velocity of 0.02125 m/s enters a pipe having an inner diameter of 2.54 cm. The pipe is 2.75 m in length, and the wall temperature is constant at 86◦ C. Determine the exit water temperature. 22.22: Water at 17◦ C at a flow rate of 2.0 kg/s enters a pipe having an inner diameter of 2.54 cm. The pipe is 5.00 m in length and the wall temperature is constant at 125◦ C. Determine the exit water temperature. 22.23: Water at 15◦ C at a flow rate of 3.12 kg/s enters a pipe having an inner diameter of 4.00 cm. The exit water temperature is to be 25◦ C and the wall temperature is constant at 90◦ C. Determine the length of the pipe. 22.24: Air at 2.5 atm and a bulk temperature of 250◦ C is heated as it flows through a tube with a diameter of 3.81 cm with a velocity of 8 m/s. The heating is via a constant heat flux and the wall temperature is 25◦ C above the air temperature. Determine the heat flux. 22.25: Water enters a pipe with an inner diameter of 5.0 cm at 10◦ C and a flow rate of 2.75 kg/s. The pipe is 4.8 m in length and the wall temperature is constant over its entire length at 70◦ C. Determine the exit water temperature.

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Introduction to Thermal and Fluid Engineering

22.26: In Problem 22.25, the wall is subjected to a constant heat flux and all other conditions remain the same. Determine the maximum tube wall temperature. 22.27: Water enters a 2.00-in-schedule 40 pipe at 7◦ C at a flow rate of 2.625 kg/s. The wall temperature is constant over the entire length of the tube at 90◦ C and the water leaves at 23◦ C. Determine the length of the pipe. 22.28: Water at 15◦ C and a flow rate of 3.12 kg/s enters a pipe with a 4-cm-inner diameter. The wall temperature is constant over the entire length of the tube at 90◦ C and the water leaves at 35◦ C. Determine the length of the pipe. The Equivalent Diameter 22.29: Determine the equivalent diameter of the trapezoidal passage shown in Figure P22.29. a

h

b FIGURE P22.29

22.30: Determine the equivalent diameter of the trapezoidal passage shown in Figure P22.30. 2a

h

2b FIGURE P22.30

22.31: Determine the equivalent diameter of the triangular passage shown in Figure P22.31.

a

b FIGURE P22.31

Forced Convection—Internal Flow

739

22.32: Determine the equivalent diameter of the isosceles right triangular passage shown in Figure P22.32.

a

a

FIGURE P22.32

22.33: Determine the equivalent diameter of the equilateral triangular passage shown in Figure P22.33.

a

a

a FIGURE P22.33

22.34: Determine the equivalent diameter of the circular passage shown in Figure P22.34.

d

FIGURE P22.34

22.35: Determine the equivalent diameter of the rectangular passage shown in Figure P22.35. b

h

FIGURE P22.35

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Introduction to Thermal and Fluid Engineering

22.36: Determine the equivalent diameter of the annular passage shown in Figure P22.36.

di

do

FIGURE P22.36

22.37: Figure P22.37 shows the interior of a circular tube containing four fins. Determine the equivalent diameter.

δ

d

b FIGURE P22.37

22.38: Determine the equivalent diameter of the finned annular passage shown in Figure P22.38.

ro ri

δ b FIGURE P22.38

Forced Convection—Internal Flow

741

22.39: Determine the equivalent diameter of the octagonal passage shown in Figure P22.39.

s All Sides Have Length, s FIGURE P22.39

Heat Transfer Coefficients in Passages Other than Circular 22.40: Lubricating oil (approximately SAE 50) flows at a velocity of 0.42 m/s in a 4-m-long rectangular duct that is 2 cm × 4 cm. The oil has a bulk temperature of 100◦ C, and the wall is at 40◦ C. Determine the heat transfer coefficient. 22.41: Lubricating oil (approximately SAE 50) flows at a velocity of 0.42 m/s in a 4-m-long equilateral triangular duct that is 2.00 cm on a side. The oil has a bulk temperature of 100◦ C, and the wall is at 40◦ C. Determine the heat transfer coefficient. 22.42: Air at 25 atm flows at a velocity of 6 m/s in a 1.25-m-long square duct that is 2.50 cm on a side. The air has a bulk temperature of 250◦ C, and the wall is at 80◦ C. Determine the heat transfer coefficient. 22.43: Air at 25 atm flows at a velocity of 6 m/s in a 1.25-m-long annular region (di = 2.50 cm, do = 4 cm). The air has a bulk temperature of 250◦ C, and the wall is at 80◦ C. Determine the heat transfer coefficient. 22.44: Rework Problem 22.43 for the case of air at 1 atm. 22.45: Nitrogen at 1 atm flows in a 1.5-m-long 2-in schedule 40 pipe that is equipped at the interior wall with four fins, each with a height of 0.75 cm and a thickness of 2.286 mm. The flow rate is 12 m/s, the bulk temperature of the nitrogen is 27◦ C, and the wall temperature is 77◦ C. Determine the heat transfer coefficient. 22.46: Nitrogen at 1 atm flows at a velocity of 12 m/s in the configuration of Figure P22.46. Pertinent dimensions are: a 1 = 4 cm

b 1 = 1.5 cm

a 2 = 3 cm

b 2 = 1.25 cm

= 40 cm The bulk temperature of the nitrogen is 27◦ C and the wall temperature is 77◦ C. Determine the heat transfer coefficient.

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Introduction to Thermal and Fluid Engineering

b2 a2 b1

a1 FIGURE P22.46

22.47: Superheated steam at atmospheric pressure and 450◦ C enters a square duct that is 20 cm on a side and 4-m long. The velocity of the steam is 1.375 m/s and the pipe wall temperature is 400◦ C. Determine the heat transfer coefficient. 22.48: Superheated steam at atmospheric pressure and 450◦ C enters a trapezoidal duct that has a height of 4 cm and bases of 20 and 24 cm and is 4-m long. The velocity of the steam is 1.375 m/s and the duct wall temperature is 400◦ C. Determine the heat transfer coefficient. 22.49: Water at a bulk temperature of 27◦ C flows in a square duct whose side dimension is 1 cm. The velocity is 3.25 m/s and the duct is 1 m in length. For a wall temperature of 47◦ C, determine the heat transfer coefficient. 22.50: Water at a bulk temperature of 27◦ C flows in the interior of an isosceles right triangular duct (45◦ -90◦ -45◦ ) whose two equal sides measure 2 cm. The velocity is 3.25 m/s and the duct is 1 m in length. For a wall temperature of 47◦ C, determine the heat transfer coefficient. 22.51: Rework Problem 22.50 for water flowing at a velocity of 0.04 m/s in a rectangular duct that is 3 cm × 4 cm. All other conditions remain the same. 22.52: Rework Problem 22.50 for ethylene glycol in a rectangular duct that is 3 cm × 4 cm. The wall temperature is 77◦ C, and the duct length is 2 m. All other conditions remain the same. 22.53: Rework Problem 22.50 for glycerin in a rectangular duct that is 3 cm × 4 cm with a wall temperature of 47◦ C and a duct length of 2 m. All other conditions remain the same. 22.54: Rework Problem 22.50 for unused engine oil in a rectangular duct that is 3 cm × 4 cm. All other conditions remain the same except for the duct wall temperature, which is at 77◦ C. 22.55: Ethylene glycol is heated from 37◦ C to 27◦ C on the inside of a circular annular region. The annular region has an inner diameter, di , of 2.50 cm and outer diameter, do , of 4.50 cm. The metal wall is held at 77◦ C, and the velocity in the annulus is 2.125 m/s. Estimate the length of the annular region to ensure that h, the heat transfer coefficient, is 1325 W/m2 -K. 22.56: Rework Problem 22.55 for an annular region resembles a pair of concentric squares. The inside wall is 2 cm on a side and the outside wall is 4 cm on a side. All other conditions remain the same.

Forced Convection—Internal Flow

743

22.57: Rework Problem 22.55 for flows through an equilateral triangular passage that is √ 2 3 cm on a side. All other conditions remain the same. 22.58: In a chilled water air-conditioning system, water enters and leaves a circular annular region (di = 3 cm and do = 4.5 cm) at 15◦ C and 5◦ C, respectively. The velocity of the water is 0.75 m/s and the annulus walls are at 2◦ C. Determine the heat transfer coefficient.

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23 Forced Convection—External Flow

Chapter Objectives To examine the factors that influence heat transfer between a solid surface and an adjacent flowing fluid. • To catalog the dimensionless empirical relations that allow the determination of the heat transfer coefficient, h, for external forced flow. • To use the appropriate empirical expressions for h in evaluating the heat transfer behavior of representative systems. •

23.1

Introduction

In this chapter, we continue to discuss and illustrate how convective heat transfer rates are determined using empirical correlations that have been generated and verified over many years of effort. Our focus will be on convective heat transfer between a solid surface and a fluid that flows over its outside surface, that is, external flow. External flows, as discussed in Chapter 16, are considerably more complex than the internal flows considered in the previous chapter. Dimensional analysis techniques were employed in Chapter 22 to generate the dimensionless parameters that are useful in determining the heat transfer coefficient, h. This same technique would yield the same parameters for external flow that were generated earlier for internal flow. We thus state, without proof, that correlations for external flow convection are expected to have the form Nu L = CRea Prb

(23.1)

where Nu L =

hL k

is the Nusselt number Re L =

ˆ L V 

is the Reynolds number and Pr =

c p k

is the Prandtl number. 745

746

Introduction to Thermal and Fluid Engineering

The subscript, L, associated with Nu L and Re L , is the length scale that is appropriate for a given geometric case. The constants, a , b, and C in Equation 23.1 have been experimentally determined for a large number of geometric and flow conditions. Subsequent sections will include discussions of several important cases. As noted earlier, physical properties of interest are temperature dependent. In heat transfer processes involving external flow, conventional practice is to evaluate properties at the film temperature, T f , which is the average of the wall and bulk fluid temperatures Tf =

23.2

Tw + T∞ 2

(23.2)

Flow Parallel to a Plane Wall

23.2.1 Laminar Boundary Layer Flow As noted in Chapter 16, the case of flow parallel to a flat surface has been solved analytically to yield operational expressions for boundary layer thickness, , and the coefficient of skin friction, C f . In Figure 23.1, the features of the hydrodynamic boundary layer are represented in a similar fashion as was illustrated in Figure 16.7. When a temperature difference exists between the plate surface and the fluid, heat transfer will occur and the fluid layers near the wall will undergo changes in temperature. This process generates a thermal boundary layer that will grow as a function of x much like the hydrodynamic boundary layer. The thickness of the thermal boundary layer is designated as t in Figure 23.1. The boundary layer region is exaggerated in Figure 23.1 for clarity; in reality it is quite thin. The fluid temperature varies from Tw to T∞ for 0 < y < t . The trends indicated in Figure 23.1 apply to the case of steady, incompressible laminar flow along a plane surface. The physical situation is one for which an analytical solution exists. This is a classic problem in the fluid mechanics literature and is referred to as the Blasius problem in recognition of the original work of H. Blasius.1 The heat transfer counterpart to the Blasius problem has also been solved and is discussed in most heat transfer texts (Welty et al., 2008 and Incropera and Dewitt, 2002). The important relationships provided by solving the governing equations of mass conservation, the first law of thermodynamics, and Newton’s second law of motion are the boundary layer thickness  = Pr1/3 (23.3) t or incorporating Equation 16.5 t = 5xRe−1/2 Pr−1/3 x

(23.4)

and the Nusselt number Nu L =

hL 1/2 = 0.664Re L Pr1/3 k

(23.5)

Equation 23.5 applies for fluids with Pr ≥ 0.6. This range includes all gases and most liquids except for the liquid metals such as mercury and molten sodium. A general 1 P.

R. H. Blasius (1883–1970) was one of Prandtl’s students who provided an analytical solution to the boundary layer equations in 1908.

Forced Convection—External Flow

747 Flow

y δt

δ x FIGURE 23.1 Laminar boundary layer growth for flow along a plane wall.

expression, which applies to any fluid in laminar flow over a flat plate, has been recommended by Churchill and Ozoe (1973): 1/2

Nu L = 

0.677Re L Pr1/3   1/4 0.0468 2/3 1+ Pr

(23.6)

Note that the thermal boundary layer thickness, t , increases as x 1/2 . The Nusselt number represented by Equation 23.6 is the mean value over the distance, L, measured from the leading edge of the plate. The reader is reminded specifically of the Reynolds number dependence of boundary layer flow along a flat plate. •

The flow is laminar for 0 < Rex < 2 × 105

•

The flow is turbulent for 3 × 106 < Rex

•

And the transition region exists for 2 × 105 < Rex < 3 × 106

Example 23.1 Air at atmospheric pressure and a temperature of 280 K flows at a velocity of 3 m/s. A plane surface, 15 cm in width at a temperature of 400 K, is oriented parallel to the direction of flow (Figure 23.2). Determine (a) the length of the plate for the boundary layer flow to remain laminar, (b) the maximum thermal boundary layer thickness for a plane surface of this length, and (c) the rate of heat transfer between the plate surface and the air. T = 400 K Air 200 K V = 3 m/s FIGURE 23.2 Configuration for Example 23.1.

15 cm

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Introduction to Thermal and Fluid Engineering

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) Radiation heat exchange with the surroundings is negligible. (3) The bottom surface of the plate is adiabatic. Fluid properties must be evaluated at the film temperature Tf =

Tw + T∞ 400 K + 280 K = = 340 K 2 2

At 340 K, the properties of interest obtained from Table A.13 in Appendix A (with interpolation) are k = 0.0292 W/m-K,

 = 1.975 × 10−5 m2 /s

and

Pr = 0.699

(a) The laminar boundary layer will exist for Re < 2 × 105 . The corresponding length is given by Re L =

L Vˆ = 2 × 105 

which yields L=

Re L  (2 × 105 )(1.975 × 10−5 m2 /s) = = 1.32 m ⇐ 3 m/s Vˆ

(b) The thermal boundary layer thickness given by Equation 23.4 will reach its maximum value at L = 1.32 m, where Rex = 2 × 105 and the value of t is determined as t = 5xRe−1/2 Pr−1/3 x = 5(1.32 m)(2 × 105 ) −1/2 (0.699) −1/3 = (6.60 m)(2.236 × 10−3 )(1.1268) = 0.0164 m

(1.64 cm) ⇐

(c) The heat transfer to a plate measuring 0.15 m × 1.32 m is determined from ˙ = h S(Tw − T∞ ) Q where h can be evaluated using Equation 23.5 hL 1/2 = 0.664Re L Pr1/3 k or h=

k 1/2 0.664Re L Pr1/3 L

Appropriate substitution yields h=

0.0292 W/m-K (0.664)(2 × 105 ) 1/2 (0.699) 1/3 1.32 m

= (0.0221 W/m2 -K)(0.664)(447.2)(0.888) = 5.85 W/m2 -K

Forced Convection—External Flow

749

and, finally, ˙ = h S(Tw − T∞ ) Q = (5.85 W/m2 -K)(0.15 m)(1.32 m)(400 K − 280 K) = 139 W

23.2.2 Turbulent Boundary Layer Flow Flow in the boundary layer will be turbulent for Rex > 3 × 106 . As mentioned in Section 16.4, the relevant fluid mechanics parameters are −1/5

 = 0.376xRe L

(16.10)

−1/5 C¯ f = 0.072Re L

(16.12)

and

The analysis leading to these relationships can be extended to include heat transfer phenomena, that is, behavior observed when a temperature difference exists between the surface and flowing fluid. The expressions that pertain to the boundary layer region are t = 0.376xRe−1/5 Pr−1/3 x

(23.7)

and Nu L =

hL L 4/5 = 0.036Re L Pr1/3 k

(23.8)

which applies for 0.60 < Pr < 60. The natural question at this point is “How should one proceed to evaluate heat transfer for flow along a plane wall when more than one type of flow is present?” A relatively simple approach to this situation is to presume two types of flow, laminar and turbulent with transition from one to the other occurring abruptly at some value of Rex lying between 2 × 105 and 3 × 106 . Quantitatively, this concept may be expressed as   xc  L 1 hL = h lam d x + h turb d x (23.9) L 0 xc In one conventional choice for xc , Re = 5 × 105 , and the result is

hL L 4/5 Nu L = = 0.036Re L − 871 Pr1/3 k

(23.10)

which applies for Rexc = 5 × 105 ,

0.60 < Pr < 60,

and

5 × 105 < Re L < 108

Example 23.2 For air flowing along a plane surface as described in Example 23.1, determine the total heat transfer over a flat plate that is 10-m long.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) Radiation heat exchange with the surroundings is negligible. (3) The bottom surface of the plate is adiabatic.

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Introduction to Thermal and Fluid Engineering

Recall that at T f = 340 K, the fluid properties of interest are  = 1.975 × 10−5 m2 /s

k = 0.0292 W/m-K,

and

Pr = 0.699

The Reynolds number, Re L , for this situation is Re L =

L Vˆ (10 m)(3 m/s) = = 1.519 × 106  1.975 × 10−5 m2 /s

Equation 23.10 applies. Solving for h, we obtain h= =

k 4/5 0.036Re L − 871 Pr1/3 L 0.0292 W/m-K  0.036(1.519 × 106 ) 4/5 − 871 (0.699) 1/3 10 m

= (0.00292 W/m2 -K)[0.036(88,154) − 871](0.8875) = 5.97 W/m2 -K The total heat transfer is then determined as ˙ = h S(Tw − T∞ ) Q = (5.97 W/m2 -K)(0.15 m)(10 m)(400 K − 280 K) = 1075 W 23.2.3 Additional Considerations 23.2.3.1 Constant Heat Flux Wall Condition The foregoing discussion and example problems presumed a constant wall temperature. As presented in Chapter 20, the other type of boundary condition of interest is that of a ˙ prescribed heating rate or wall heat flux, q˙ = Q/S. When the wall heat flux is prescribed, the total heat transfer is easily calculated as ˙ = q˙ S Q

(23.11)

The other quantity of interest is the wall temperature that is related to the heat flux according to ˙ = h S(Tw − T∞ ) Q or Tw = T∞ +

˙ Q/S h

or Tw = T∞ +

˙ Q/S (k/L)Nu L

(23.12)

The evaluation of Nu L for a constant wall heat flux is discussed in considerable detail by Kays and Crawford (1993). They conclude that for external flow, the Nusselt numbers for constant heat rate and constant wall temperature differ, quantitatively, by approximately 2%. In light of other inaccuracies in determining h it is concluded that Equation 23.12 may be used with any of the constant surface temperature expressions to determine the wall temperature.

Forced Convection—External Flow

751

Flow y

δt

δ x X FIGURE 23.3 Two boundary layers with an unheated starting length.

23.2.3.2 Unheated Starting Length All cases considered thus far presume that the wall surface is heated over its entire length. Figure 23.3 shows the two boundary layers for the case of a hydrodynamic boundary that begins at the leading edge, x = 0, but with the thermal boundary layer at some starting length, X, downstream from the leading edge. The wall boundary condition can be expressed as T = T∞

0 0.700 and Ra < 1011   −4/9 hd 0.469 9/16 1/4 1+ = 2.00 + 0.589 Ra (24.12) k Pr where all properties except  are obtained at the film temperature. The value of  for gases is obtained at the temperature of the surroundings; for liquids, it is obtained at the film temperature. The range for the Prandtl number is 0 < Pr < ∞. 24.3.6 Horizontal Plates For the vertical plate that has its significant dimension in line with the gravity vector, the natural convection currents flow parallel to the surface. However, for the horizontal plate, the significant dimension, whatever its magnitude, is normal to the direction of the flow, and we find a vast difference in the natural convection current patterns. For example, with a horizontal heated plate facing downward, we see that to escape into the free stream, the dense air at the bottom surface must flow laterally to the edges of the plate. On the other hand, if the heated surface faces upward, the less dense air tends to rise but is impeded by the ever-present cooler air that flows downward. In these cases, the correlations of McAdams (1954), Goldstein et al. (1973), and Lloyd and Moran (1974) are usually employed. These correlations are for hot plates facing upward or cold plates facing downward hL 1/4 = 0.54 Ra L k

(2.6 × 104 < Ra L < 3 × 107 )

(24.13a)

(107 < Ra L < 3 × 1010 )

(24.13b)

and hL 1/3 = 0.15 Ra L k

and for hot plates facing downward and cold plates facing upward hL 1/4 = 0.27 Ra L k

(3 × 105 < Ra L < 3 × 1010 )

(24.13c)

In Equations 24.13, the significant dimension for the Rayleigh number is L=

plate area plate perimeter

(24.14)

Example 24.4 A 0.50-m2 plate with a surface temperature 120◦ C is suspended horizontally in a vat of still water at 20◦ C (Figure 24.6). Determine the heat loss from both faces of the plate.

Solution Assumptions and Specifications (1) Steady state exists. (2) Buoyancy effects are included in the correlation employed. (3) The ambient fluid is quiescent.

Free or Natural Convection

785 50 cm Ts = 120°C

50 cm Still Water at T∞ = 20°C FIGURE 24.6 Configuration for Example 24.4.

(4) Radiation effects are negligible. (5) The plate dissipates from both faces. (6) The acceleration of gravity is taken as 9.81 m/s2 . Still water and a horizontal surface are specified. Here we use Equation 24.14 for the significant dimension. L=

plate area L2 L 0.50 m = = = = 0.125 m plate perimeter 4L 4 4

In this case, the temperature difference will be T = Ts − T∞ = 120◦ C − 20◦ C = 100◦ C and the film temperature is Tf =

Ts + T∞ 120◦ C + 20◦ C = = 70◦ C 2 2

and Table A.14 can be used to find k = 0.664 W/m-K

 = 0.4173 × 10−6 m/s2

 = 0.590 × 10−3 K−1

Pr = 2.573

The Rayleigh number is determined with L = 0.125 m as the significant dimension:   gL 3 T Ra = Gr Pr = Pr 2 =

(9.81 m/s2 )(0.590 × 10−3 K−1 )(0.125 m) 3 (100◦ C)(2.573) (0.4173 × 10−6 m/s2 ) 2

= 1.670 × 1010 We can use Equation 24.13b for the top surface and Equation 24.13c for the bottom surface. For the top surface Nu =

hL = 0.15 Ra1/3 k

= 0.15(1.670 × 1010 ) 1/3 = 0.15(2556.4) = 383.4

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Introduction to Thermal and Fluid Engineering

and for the bottom surface Nu =

hL = 0.27 Ra1/4 k

= 0.27(1.670 × 1010 ) 1/4 = 0.27(359.5) = 97.06 These values of the Nusselt number yield the heat transfer coefficients. For the top     k 0.664 W/m-K h = 383.6 = 383.6 = 2036.7 W/m2 -K L 0.125 m and for the bottom

    k 0.664 W/m-K h = 97.06 = 97.06 = 515.6 W/m2 -K L 0.125 m

The surface area for each face of the plate is S = L 2 = (0.50 m) 2 = 0.25 m2 Thus, the rates of heat transfer will be ˙ = h ST = (2036.7 W/m2 -K)(0.25 m2 )(100◦ C) = 50.92 kW ⇐ Q for the top and ˙ = h ST = (515.6 W/m2 -K)(0.25 m2 )(100◦ C) = 12.89 kW ⇐ Q for the bottom. We note that the dissipation from the top of the plate is about four times the dissipation from the bottom.

24.4

Natural Convection in Parallel Plate Channels

24.4.1 The Elenbaas Correlation Elenbaas (1942) was apparently the first to document a detailed study of the thermofluid behavior of a vertical channel in natural convection. His experimental results for isothermal plates in air were later confirmed by others. A unified picture of the thermal transport in such channels emerged from these and complementary studies. Figure 24.7 shows a gap between two isothermal plates. Individual thermal and velocity boundary layers are in evidence along each surface, and the rates of heat transfer approach those associated with laminar flow along isolated plates in infinite media. The development of the resulting boundary layers progresses toward the fully developed condition at the channel outlet. Alternatively, for long narrow channels, the boundary layers merge near the entrance and we observe that fully developed flow prevails along much of the channel. The analytic Nusselt number relations for the fully developed flow region and the isolated plate region can be expected to bound the Nusselt number values over the complete range of flow development. In the fully developed limit and for an isothermal channel, in the

Channel Outlet

Free or Natural Convection

787

Twall

Mid-Channel

Twall

Channel Inlet

Twall

Tambient

Temperature Profile

Velocity Profile

(a)

(b)

FIGURE 24.7 Developing temperature and velocity profiles for air in natural convection in a vertical channel: (a) the temperature profiles and (b) the velocity profiles. Broken lines designate isothermal channel boundaries.

fully developed limit, the gap-based Nusselt number Nuz ≡

hz k

is seen to depend on a gap-based Rayleigh number with z taken as the spacing between the plates Raz ≡

gz4 T Pr L2

In recognition of his seminal contributions to the field of natural convection in rectangular channels, the gap-based Rayleigh number is referred to as the Elenbaas number, El. Thus, El = Raz =

2 gc p z4 T = z4 Lk

(24.15)

2 gc p T Lk

(24.16)

where  is a thermal property parameter =

Elenbaas used his understanding of these bounding relationships to establish the form of the relationship between the Elenbaas number and the fluid properties over the entire range of fluid properties. Application of this form to the data for relatively short isothermal plates in the air yielded the Elenbaas correlation Nuz =

hz El = (1 − e −35/El ) 3/4 k 24

(24.17)

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Introduction to Thermal and Fluid Engineering

Isolated plate limit

10 5 2 1 Nu

0.5 0.2

Composite relation Fully developed limit

0.1

2×102

00.5

103

5×103

5×104

00.2 00.1 0.005 0.1 0.2 0.5 1 2

5 10 20 50 El = (Gr.Pr z/L)

FIGURE 24.8 Nusselt number variation for symmetric isothermal plates. (Data points from Elenbaas, 1942.)

24.4.2 A Composite Relation Churchill and Usagi (1972) have shown that when a function is known to vary smoothly between two limiting expressions that are, themselves, well defined and when solutions for intermediate values are difficult to obtain, an approximate composite relationship can be obtained by summing the two limiting expressions. Based on the Churchill-Usagi procedure, Bar-Cohen and Rohsenow (1984) provided a composite equation for isothermal plates   576 2.87 −1/2 Nuz = + (24.18) ( El) 2 ( El) 1/2 where Nuz and El (including ) are based on properties evaluated at the average film temperature. The validity of this solution is borne out by the close proximity of the Elenbaas (1942) data points to the composite relationship of Equation 24.18 and the asymptotic equations at both limits as indicated in Figure 24.8.

Example 24.5 Determine the heat transfer coefficient for air in natural convection within a 15-cm-long vertical parallel plate channel (Figure 24.9). The ambient air temperature is 23◦ C, the channel walls are maintained at 77◦ C, and plate spacings of (a) 2 cm, (b) 0.50 cm, and (c) 0.20 cm are to be considered.

Solution Assumptions and Specifications (1) Steady state exists. (2) The air may be treated as an ideal gas. (3) Buoyancy effects are included in the correlation employed. (4) The ambient fluid is quiescent. (5) Radiation effects are negligible. (6) The acceleration of gravity is taken as 9.81 m/s2 .

Free or Natural Convection

789

L = 15 cm

z z

FIGURE 24.9 Configuration for Example 24.5.

We will calculate the Elenbaas number as a function of the plate spacing for all three spacings first. Here T = Ts − T∞ = 77◦ C − 23◦ C = 54◦ C and the film temperature is Tf =

Ts + T∞ 77◦ C + 23◦ C = = 50◦ C 2 2

(323 K)

Then with =

1 = 3.096 × 10−3 K−1 323 K

Table A.13 provides (with interpolation) k = 0.0279 W/m-K

 = 19.52 × 10−6 kg/m-s

 = 1.0949 kg/m3

c p = 1007.2 J/kg-K

Here =

2 gc p T k L

= (1.0949 kg/m3 ) 2 (9.81 m/s2 ) ×

(3.096 × 10−3 K−1 )(1007.2 J/kg-K)(54◦ C) (19.52 × 10−6 kg/m-s)(0.0279 W/m-K)(0.15 m)

= 2.424 × 1010 m−4

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Introduction to Thermal and Fluid Engineering

and El = z4 = (2.424 × 1010 m−4 )z4 (a) For a plate spacing of 2 cm El = (2.424 × 1010 m−4 )(0.02 m) 4 = 3880 and

 Nuz =

576 2.87 + 2 ( El) ( El) 1/2

−1/2

 =

576 2.87 + 2 (3880) (3880) 1/2

−1/2

= (3.826 × 10−5 + 0.0461) −1/2 = 4.656 Then h = Nuz

    k 0.0279 W/m-K = 4.656 = 6.50 W/m2 -K ⇐ z 0.02 m

(b) For a plate spacing of 0.50 cm El = (2.424 × 1010 m−4 )(0.005 m) 4 = 15.15 and

 Nuz =

576 2.87 + 2 ( El) ( El) 1/2

−1/2

 =

576 2.87 + 2 (15.15) (15.15) 1/2

−1/2

= (2.5096 + 0.7374) −1/2 = 0.5550 Then h = Nuz

    k 0.0279 W/m-K = 0.5550 = 3.10 W/m2 -K ⇐ z 0.005 m

(c) For a plate spacing of 0.20 cm El = (2.424 × 1010 m−4 )(0.002 m) 4 = 0.3880 and

 Nuz =

576 2.87 + 2 ( El) ( El) 1/2

−1/2

 =

576 2.87 + 2 (0.3880) (0.3880) 1/2

−1/2

= (3826.1 + 4.608) −1/2 = 0.0162 Then h = Nuz

    k 0.0279 W/m-K = 0.0162 = 0.225 W/m2 -K ⇐ z 0.002 m

24.4.3 Optimum Plate Spacing An examination of Figure 24.8 shows that the rate of heat transfer from each plate decreases as the plate spacing, z, is decreased. Because the total number of plates or the plate surface area increases with reduced spacing, we see that the total heat transfer may be maximized by finding the plate spacing at which the total plate surface-heat transfer coefficient product

Free or Natural Convection

791

is maximized. Bar-Cohen and Rohsenow (1984) point out that Elenbaas (1942), based on his experimental results, determined that this optimum spacing for negligibly thick plates could be obtained by setting El = 46 leading to zopt =

2.714 1/4

(24.19)

where the thermal property parameter, , is defined in Equation 24.16. We now turn to Design Example 16, which is a rather comprehensive problem involving the design of a heat sink for a package of electronic equipment.

24.5

Design Example 16

A vertical surface, W = 4 cm by L = 4 cm (Figure 24.10) must dissipate 2.25 W at 75◦ C by natural convection to the air at 45◦ C. Determine whether the surface without fins is sufficient to accommodate the required dissipation. Determine the number of fins with a height b = 3 cm, a length L = 4 cm, a thickness, = 2.286 mm, and k = 180 W/m-K that will be required if the bare surface fails to provide the required dissipation.

Solution Assumptions and Specifications (1) Steady state exists. (2) The air may be treated as an ideal gas. (3) Buoyancy effects are included in the correlation employed. (4) The ambient fluid is quiescent. (5) Radiation effects are negligible. (6) The acceleration of gravity is taken as 9.81 m/s2 . The configuration is shown in Figure 24.10. In order to make sure that the fins, if used, are isothermal plates, we will aim for a fin efficiency of at least 0.990, thereby ensuring a negligible temperature drop from base to tip. Here, the film temperature is Tf =

Ts + T∞ 75◦ C + 45◦ C = = 60◦ C 2 2

(333 K)

and T = Ts − T∞ = 75◦ C − 45◦ C = 30◦ C Then at T∞ = 45◦ C

(318 K) =

1 = 3.148 × 10−3 K−1 318 K

and Table A.13 provides (with interpolation) k = 0.0287 W/m-K,

 = 19.04 × 10−6 m/s2 ,

and

Pr = 0.701

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Introduction to Thermal and Fluid Engineering

4 cm

4 cm

(a)

4 cm 0.2286 cm

4 cm

3 cm (b) FIGURE 24.10 Configuration for Design Example 16: (a) a vertical plate and (b) the vertical plate equipped with longitudinal fins of rectangular profile.

The Rayleigh number is  Ra = Gr Pr = =

gL 3 T 2

 Pr

(9.81 m/s2 )(3.148 × 10−3 K−1 )(0.04 m) 3 (30◦ C)(0.701) (19.04 × 10−6 m/s2 ) 2

= 1.147 × 105 We employ Equation 24.7a, which is applicable for the case of the vertical plate without the fins. Nu = 0.55 Ra1/4 = 0.55(1.147 × 105 ) 1/4 = 0.55(18.40) = 10.12 Then h = Nu

    k 0.0287 W/m-K = 10.12 = 7.26 W/m2 -K ⇐ L 0.04 m

Free or Natural Convection

793

and with S = (0.04 m) 2 = 0.0016 m2 we obtain the heat dissipation as ˙ = h ST = (7.26 W/m2 -K)(0.0016 m2 )(30◦ C) = 0.349 W < 2.25 W Q We will need to use the fins and when they are employed at the film temperature of 60◦ C (333 K), Table A.13 provides (with interpolation) k = 0.0287 W/m-K

 = 19.97 × 10−6 m/s2

 = 1.059 kg/m3

c p = 1007.9 J/kg-K

and with  taken at the film temperature =

1 = 3.003 × 10−3 K−1 333 K

Here =

2 gc p T k L

= (1.059 kg/m3 ) 2 (9.81 m/s2 ) ×

(3.003 × 10−3 K−1 )(1007.9 J/kg-K)(30◦ C) (19.97 × 10−6 m/s2 )(0, 0287 W/m-K)(0.04 m)

= 4.357 × 1010 m−4 The optimum clear space between the fins will then be zopt =

2.714 2.714 = = 5.940 × 10−3 m 1/4  456.89 m1

(0.594 cm)

and the number of fins required is nf =

W 4.00 cm 4.00 cm = = = 4.86 fins zopt + 0.5940 cm + 0.2286 cm 0.8226 cm

We will call for five fins with the knowledge that the fin spacing will be slightly smaller than optimum. Moreover, we will space the fins so that there will be five equal spaces z+ =

4.00 cm = 0.8000 cm 5 fins

so that z = 0.8000 cm − = 0.8000 cm − 0.2286 cm = 0.5714 cm The Elenbaas number will be El = z4 = (4.357 × 1010 m−4 )(5.714 × 10−3 m) 4 = 46.45

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Introduction to Thermal and Fluid Engineering

and the Nusselt number will then be    −1/2 576 2.87 −1/2 576 2.87 Nuz = + = + ( El) 2 ( El) 1/2 (46.45) 2 (46.45) 1/2 = (0.2670 + 0.4211) −1/2 = 1.2055 Then

    k 0.0287 W/m-K h = Nu = 1.2055 = 6.06 W/m2 -K z 5.714 × 10−3 m

Before we make the surface computation, we will check to see if the fin efficiency is close to 0.990. Following the procedure illustrated in Section 20.7.3 with L  , we have  m=

2h k



1/2 =

2(6.06 W/m2 -K) (180 W/m-K)(2.286 × 10−3 m)

1/2 = 5.427 m−1

and mb = (5.427 m−1 )(0.03 m) = 0.1628 The fin efficiency is

=

tanh mb tanh 0.1628 = = 0.991 mb 0.1628

This checks the assumption of the fin efficiency and allows us to accept the foregoing calculations. The surface area of the base plate will be the area, WL , minus the footprints of the five fins: Sb = L(W − n f ) = (4.00 cm)[4.00 cm − 5(0.2286 cm)] = (4.00 cm)[4.00 cm − 1.143 cm) = (4.00 cm)(2.857 cm) = 11.428 cm2 The surface area of the five fins is S f = 2n f b L = 2(5)(3.00 cm)(4.00 cm) = 120.0 cm2 Thus, the total surface area for this application is S = Sb + S f = 11.428 cm2 + 0.991(120 cm2 ) = 11.428 cm2 + 118.92 cm2 = 130.35 cm2 and the heat transferred will be ˙ = h ST = (6.06 W/m2 -K)(0.0130 m2 )(30◦ C) = 2.36 W > 2.25 W ⇐ Q This design is deemed to be satisfactory.

Free or Natural Convection

795

TH

TC

. Q

H

L FIGURE 24.11 A pair of rectangular plates that form an enclosure oriented in the vertical direction.

24.6

Natural Convection in Enclosures

The cavity or enclosure shown in Figure 24.11 has a height, H, and a plate spacing, L . Here, the two opposing plates are held at different temperatures, TH and TC , where TH > TC . The dimension perpendicular to the plane of the paper, W, is presumed to be very large, that is W  L. This configuration is of more than casual interest. For example, the air spaces between two vertical panes of glass or between two vertical walls in a structure are two important examples of such an enclosure. Moreover, the air space enclosed by the absorbing surfaces of a solar collector and its cover plate (glass or plastic) provides us with another important example. In the case of the solar collector, the enclosure is usually tilted at some angle, , from the vertical. When considering enclosures, we define the aspect ratio, , as =

H L

(24.20)

the Nusselt number with respect to L hL k

(24.21)

g(TH − TC )L 3 2

(24.22)

Nu L = the Grashof number with respect to L Gr L =

and the Rayleigh number with respect to L Ra L = Gr L Pr =

g(TH − TC )L 3 Pr 2

(24.23)

In Equations 24.22 and 24.23,  is evaluated at the average of the two plate temperatures.

796

Introduction to Thermal and Fluid Engineering

In the limiting case where there are no convection currents, heat transfer between the hot wall, at T = TH , and the cooler one, at T = TC , will be by conduction only. In such a case h=

k L

and, consequently, Nu L = 1 This lower limit for Nu L is often referred to as the conduction limit. The effective or apparent conductivity for an enclosure gap, defined as ke ≡ Nu L k

(24.24)

allows the rate of heat transfer to be expressed, using the Fourier law, in the form ˙ = ke S(TH − TC ) Q L

(24.25)

24.6.1 Working Correlations 24.6.1.1 Vertical Rectangular Enclosures For conditions where 2 <  ≤ 10, Pr < 105 , and 105 < Ra L < 1010 , Catton (1978) recommends   Pr Ra L 0.28 −1/4 Nu L = 0.22  (24.26) 0.22 + Pr When 1 <  < 2, 103 < Pr < 105 , and the quantity in the parentheses is greater than 103 , the recommended expression is  Nu L = 0.18

Pr Ra L 0.22 + Pr

0.29 (24.27)

For larger aspect ratios, McGregor and Emory (1969) proposed Nu L = 0.42 Ra L Pr0.012 −0.30 1/4

(24.28)

for 10 <  < 40, 1 < Pr < 2 × 106 , and 104 < Ra L < 107 , and 1/3

Nu L = 0.046 Ra L

(24.29)

when 1 <  < 40, 1 < Pr < 20, and 106 < Ra L < 109 . 24.6.1.2 Tilted Vertical Enclosures Figure 24.12 shows an inclined vertical enclosure with the angle  measured from the horizontal. Correlations are available for 0◦ <  < 70◦ , but we present the correlation for 0◦ <  < 70◦ for the air-filled enclosure with  < 12 and for 0 < Ra L < 105 (Hollands et al., 1970). The maximum allowable value of the inclination angle, , is a function of the aspect ratio. Table 24.1 provides these values. Nu L = 1.00 + 1.441 2 + 3

(24.30)

Free or Natural Convection

797

L

H

θ

FIGURE 24.12 A tilted enclosure.

where 1 = 1 −

1708 Ra L cos 

2 = 1 −

1708(sin 1.8) 1.60 Ra L cos 

and  3 =

Ra L cos  5830

1/3 −1

If either 1 or 2 is negative, its value is set equal to zero. 24.6.2 Concentric Cylinders For the case of the concentric horizontal cylinder arrangement shown in Figure 24.13a, the correlation of Raithby and Hollands (1975) provides the effective gap thermal conductivity, ke , to be used in ˙ Q 2ke = (To − Ti ) L ln do /di

(24.31)

TABLE 24.1

Maximum Allowable Value of the Angle of Inclination,  Λ = H/L



1 3 6 12 >12

25◦ 53◦ 60◦ 67◦ 70◦

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Introduction to Thermal and Fluid Engineering

di

di

do

L

(a)

(b)

FIGURE 24.13 The gap formed by (a) concentric cylinders and (b) concentric spheres.

where the subscripts i and o refer, respectively, to the inner and outer cylinder. Here  ke = 0.386k

Pr 0.861 + Pr

1/4 1/4

(24.32)

where 

=

do ln di −3/5

4 Ra L −3/5 5 )

L 3 (do

+ di

L=

do − di 2

and

is the gap width. Equation 24.32 may be used in the range 102 <  < 107 and if  < 100, ke is taken as ke = k. Otherwise, ke must be greater than k. 24.6.3 Concentric Spheres Raithby and Hollands (1975) also considered the case of concentric spheres and proposed that ˙ = ke  do di (To − Ti ) Q L

(24.33)

with ke expressed as  ke = 0.74k

Pr 0.861 + Pr

1/4 1/4

and =

L Ra L −7/5 (do di ) 4 (do

−7/5 5 )

+ di

Equation 24.34 is valid in the range 102 <  < 107 .

(24.34)

Free or Natural Convection

24.7

799

Summary

The reader is cautioned that the following summary is simply a guide to the working correlations. Reference must always be made to the conditions of applicability, described in the text, that pertain to these correlations. The dimensionless correlations to be employed in design and analysis of configurations involving natural convection heat transfer are based on the Grashof and Rayleigh numbers. Gr ≡

2 gL 3 T 2

(24.4a)

and Ra = Gr Pr The working correlations provided in the text are summarized in the following tabulation: Configuration

Equation(s)

Vertical plates Inclined vertical plates Vertical cylinders Horizontal cylinders Spheres Hot plates facing upward Hot plates facing downward Vertical rectangular enclosures Concentric cylinders Concentric spheres

24.7 and 24.8 24.8a 24.9 24.10 and 24.11 24.12 24.13a and 24.13b 24.13c 24.26 to 24.29 24.31 24.33

For parallel plate channels, the Elenbaas number is used with z as the gap spacing El = Raz =

gz4 T Pr L2

and the Nusselt number is obtained from Equation 24.18.

24.8

Problems

Vertical Plates 24.1: Still water at 15◦ C is heated on both sides of an 18-cm × 18-cm-vertical plate held at 95◦ C. Use the McAdams correlation to predict the heat transferred. 24.2: Still water at 15◦ C is heated on both sides of an 18-cm × 18-cm-vertical plate held at 95◦ C. Use the Churchill-Chu correlation to predict the heat transferred. 24.3: Still air at atmospheric pressure and 15◦ C is heated on both sides of an 18-cm × 18-cm-vertical plate held at 95◦ C. Use the McAdams correlation to predict the heat transferred. 24.4: Still air at atmospheric pressure and 15◦ C is heated on both sides of an 18-cm × 18-cm-vertical plate held at 95◦ C. Use the Churchill-Chu correlation to predict the heat transferred.

800

Introduction to Thermal and Fluid Engineering

24.5: Still air at 2 atm and 15◦ C is heated on both sides of an 18 cm × 18 cm vertical plate held at 95◦ C. Use the McAdams correlation to predict the heat transferred. 24.6: One side of a vertical plate that is 12-cm wide and 15-cm high is held at atmospheric pressure and a temperaure of 82◦ C in a still-air environment of 22◦ C. The emissivity of the plate is 0.76, the reverse side of the plate is insulated, and the enclosure is also at 22◦ C. Use the Churchill-Chu correlation to determine the heat transferred. 24.7: Suppose that in Problem 24.6, the heat loss is 22 W. Using the McAdams correlation, determine the surface temperature to the nearest degree if all other conditions remain the same. 24.8: Use the McAdams correlation to estimate the heat loss by natural convection from a vertical wall 2-m high and 3-m wide. At 37◦ C, the wall is exposed to a still-nitrogen environment at 1 atm at 17◦ C. 24.9: Use the Churchill-Chu correlation to estimate the heat loss by natural convection from a vertical wall 2-m high and 3-m wide held at 37◦ C. The wall is exposed to a still-nitrogen environment at 1 atm at 17◦ C. 24.10: Still air at 0.94 atm and 17◦ C receives heat by natural convection from the door of a dishwasher that is held at 37◦ C and is 60-cm wide and 80-cm high. Determine the heat loss using the McAdams correlation. 24.11: For a vertical plate at 1 atm and 94◦ C in a still-air environment at 60◦ C, determine the plate height needed to make the Rayleigh number equal to 109 . 24.12: For a vertical plate at 0.15 atm and 94◦ C in a still-air environment at 60◦ C, determine the plate height needed to make the Rayleigh number equal to 109 .

Vertical Cylinders and Tilted Vertical Plates 24.13: A vertical cylinder has a diameter of 8 cm and a height of 28 cm. The cylinder is held at 85◦ C in a still-air environment at atmospheric pressure and at 25◦ C. Neglect the heat transferred from the top of the cylinder and determine the heat dissipation. 24.14: In Problem 24.13, determine the minimum diameter that will permit the cylinder to be treated as a vertical plate. 24.15: A vertical cylinder that is 1.75-m high and 20 cm in diameter is held at 97◦ C. It is used as a heater in an unstirred tank of water at 23◦ C. The top of the cylinder is insulated. Determine whether the cylinder may be treated as a vertical plate and, if so, use the Churchill-Chu correlation to establish the rate of heat transfer. 24.16: A vertical cylinder that is 2-m high and 25 cm in diameter is held at 114◦ C. The cylinder is used in stagnant air at 40◦ C and 1 atm. Neglect heat transfer from the top of the cylinder and determine the rate of heat transfer to the air (a) without radiation and (b) with radiation if the emissivity of the surface of the cylinder is 0.825. 24.17: A vertical cylinder is 80-cm high. It is used in a large tank to heat glycerin held at a bulk temperature of 13◦ C. The top of the cylinder does not dissipate. Determine the heat transferred using the McAdams correlation. 24.18: Assume that an adult male can be approximated by a vertical cylinder that is 1.95-m high and 32 cm in diameter. With normal body temperature at 37◦ C and a still-air environment at 20◦ C, determine the heat transferred using the McAdams correlation. 24.19: The vertical plate in Problem 24.1 is tilted at an angle of 38◦ from the vertical, and all other conditions remain the same, use the McAdams correlation to predict the heat transferred.

Free or Natural Convection

801

24.20: The vertical plate in Problem 24.4 is tilted at an angle of 40◦ from the vertical and all other conditions remain the same, use the Churchill-Chu correlation to predict the heat transferred. 24.21: A 25-cm × 25-cm-vertical plate is held at 74◦ C and is inclined at an angle of 56◦ from the vertical. Use the McAdams correlation to predict the heat loss to a still-air environment at 0.625 atm and 20◦ C. 24.22: The plate in Problem 24.21 is tilted at an angle of 50◦ from the vertical and is exposed to oxygen at 20◦ C at 0.75 atm. Determine the heat loss. Horizontal Cylinders 24.23: A 2.5-cm-diameter cylinder is placed horizontally in a pool of water at 20◦ C. The surface of the cylinder is maintained at 60◦ C and the cylinder is 1-m long. Use the McAdams correlation to predict the heat dissipated by the cylinder. 24.24: A 2.5-cm-diameter cylinder is placed horizontally in a pool of water at 20◦ C. The surface of the cylinder is maintained at 60◦ C and the cylinder is 1-m long. Use the Churchill-Chu correlation to predict the heat dissipated by the cylinder. 24.25: A 2.5-cm-diameter cylinder is placed horizontally in still air 1 atm and 20◦ C. The surface of the cylinder is maintained at 60◦ C and the cylinder is 1-m long. Use the McAdams correlation to predict the heat dissipated by the cylinder. 24.26: A 2.5-cm-diameter cylinder is placed horizontally in still air at 1 atm and 20◦ C. The surface of the cylinder is maintained at 60◦ C and the cylinder is 1-m long. Use the Churchill-Chu correlation to predict the heat dissipated by the cylinder. 24.27: A horizontal cylindrical heater with a diameter of 2.5 cm and a surface temperature of 117◦ C is placed in a pool of ethylene glycol at 37◦ C. For a length of 40 cm, determine the heat transferred using the McAdams correlation. 24.28: A horizontal cylindrical heater with a diameter of 2.5 cm and a surface temperature of 117◦ C is placed in a pool of ethylene glycol at 37◦ C. For a length of 40 cm, determine the heat transferred using the Churchill-Chu correlation. 24.29: A horizontal rod with a diameter of 1.75 cm is to be held at 129◦ C in still air at 1 atm and 25◦ C. The rod is 5-m long and has a resistivity of 180 -cm2 /cm. Neglect radiation and using the McAdams correlation, determine the current required. 24.30: A horizontal rod with a diameter of 1.75 cm is to be held at 129◦ C in still air at 1 atm and 25◦ C. The rod is 5-m long, has a resistivity of 180 -cm2 /cm, and an emissivity of 0.845. Assume that the radiation is to surroundings at 25◦ C, and using the McAdams correlation, determine the current required. 24.31: A horizontal steam pipe has a diameter of 6 cm and a surface temperature of 127◦ C. The pipe passes through a room at atmospheric pressure with walls and still air at 27◦ C. The pipe is 2-m long and the surface emissivity is 0.555. Using the McAdams correlation, determine the pipe heat dissipation. 24.32: A horizontal steam pipe has a diameter of 6 cm and a surface temperature of 127◦ C. The pipe passes through a room where the pressure is 0.80 atm with walls and still air at 27◦ C. The pipe is 2-m long and the surface emissivity is 0.555. Using the McAdams correlation, determine the pipe heat dissipation. Spheres 24.33: A 5-cm-diameter sphere is suspended in still air at one atmosphere and 20◦ C. The surface of the sphere is at 80◦ C. Determine the heat loss.

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Introduction to Thermal and Fluid Engineering

24.34: A 5-cm-diameter sphere is suspended in water at 20◦ C and the surface of the sphere is at 80◦ C. Determine the heat loss. 24.35: The skin of a spherical gondola of a high-altitude balloon with a diameter of 4 m is to be held at 27◦ C in a still-air environment at 1 atm and 17◦ C. Neglecting radiation, determine the heat loss from the gondola. 24.36: The skin of a spherical gondola of a high-altitude balloon with a diameter of 4 m is to be held at 23◦ C in a still-air environment at 0.20 atm and 17◦ C. Neglecting radiation, determine the heat loss from the gondola. 24.37: A sphere that is 10 cm in diameter with a skin temperature of 127◦ C is placed in still-carbon dioxide at 1 atm and 27◦ C. Neglecting radiation, determine the heat loss from the sphere. 24.38: Suppose that the carbon dioxide in Problem 24.37 is at 5 atm, but all other conditions remain the same. Determine the heat loss from the sphere. 24.39: Determine the fraction of the dissipation that is due to radiation if the emissivity of the sphere in Problem 24.37 is 0.94. 24.40: Determine the diameter of the sphere necessary to raise the natural convection heat dissipation to 30 W if all other conditions in Problem 24.37 remain the same. 24.41: Determine the heat loss by natural convection if the sphere in Problem 24.37 is placed in a pool of ethylene glycol with all other conditions remaining the same. 24.42: Determine the heat loss by natural convection if the sphere in Problem 24.37 is placed in a pool of water with all other conditions remaining the same. Horizontal Plates 24.43: A horizontal plate in the form of an isosceles right triangle whose equal sides are each 32-cm long is held at 70◦ C in still air at atmospheric pressure and 20◦ C. Determine the heat loss by natural convection from the upper surface of the plate. 24.44: A horizontal plate in the form of an isosceles right triangle whose equal sides are each 32-cm long is held at 70◦ C in water at 20◦ C. Determine the heat loss by natural convection from the upper surface of the plate. 24.45: The plate in Problem 24.3 is placed horizontally in a still-air environment at two atmopheres with all other conditions remaining the same. Determine the heat loss by natural convection from the upper surface of the plate. 24.46: The plate in Problem 24.3 is in a horizontal position at 1 atm and all other conditions remain the same. Determine the heat loss by natural convection from the upper surface of the plate. 24.47: A cast-iron plate is 28 cm × 36 cm is used as a grill in a restaurant. It is held horizontally at 82◦ C and sits in a still-air environment at 1 atm and 22◦ C. The emissivity of the cast-iron is 0.82 and the walls of the restaurant are at 17◦ C. Determine the total heat dissipation of the grill. 24.48: The top of a disk 40 cm in diameter is used to heat unused engine oil (SAE 50) that is at 4◦ C. The disk is placed face up at the bottom of the engine oil container. Determine the power required to maintain the surface temperature of the disk at 70◦ C. 24.49: A rectangular duct of height 40 cm and width 1 m carries conditioned air such that its surface temperature is 20◦ C. It is placed in a confined space containing still air at 70◦ C such that its length is 4 m. Assuming that the airflow is such that the 20◦ C

Free or Natural Convection

803

temperature is maintained, determine the total natural convection heat gained for the duct. 24.50: A steel sheet that is 3 m × 3 m is removed from a furnace where the temperature is 400◦ C. It is placed in a still-air environment at 1 atm and 20◦ C. Taking 0.28 as the emissivity of the steel and considering the sheet as a horizontal surface facing upward, determine the rate of heat loss from the surface. 24.51: A cube that is 24 cm on a side is suspended by nonconducting strings in still air at 1 atm and 40◦ C. The surfaces of the cube are held at 94◦ C, radiation is to be neglected, and the top surface of the cube is absolutely horizontal. Determine the heat dissipation from the cube surface. 24.52: A cube that is 24 cm on a side is suspended by nonconducting strings in water at 40◦ C. The surfaces of the cube are held at 94◦ C, radiation is to be neglected, and the top surface of the cube is absolutely horizontal. Determine the heat dissipation from the cube surface. Arrays of Vertical Plates and Fins 24.53: Two plates that are each 16 cm × 16 cm are located with a clear space of 1 cm between them in still air at 1 atm and 17◦ C. Each plate is at a temperature of 77◦ C. Determine the heat transfer by natural convection from the plates to the air. 24.54: Consider an aluminum (k = 182 W/m-K) longitudinal fin of uniform thickness with an adiabatic tip 16-cm high, 2.5-cm thick, and 50 cm in vertical length. The fin base temperature Tb is 120◦ C and the still-air environment at 1 atm is at 20◦ C. Consider the length dimension as the base of the fin and determine the rate of heat dissipation by natural convection from the surface of the fin. 24.55: Twenty-four vertical aluminum (k = 200 W/m-K) longitudinal fins with adiabatic tips, each with a height of 5 cm, a thickness of 2.286 mm, and a vertical length of 16 cm are placed over the rear wall of an electronic chassis that is 25-cm wide. Each fin base temperature, Tb , is 85◦ C, and the still-air environment at 1 atm is at 25◦ C. Determine the heat dissipation by natural convection by the fins. 24.56: Suppose that in Problem 24.55, it is required that the heat dissipation from the finned arrangement is 145 W. Determine how this can be accomplished if the base temperature of each fin remains at 85◦ C. 24.57: In Problem 24.53, make an estimate of the optimum spacing between the fins and determine, if, and by how much, the heat dissipation capability improves. 24.58: A vertical plate at the rear of an electronic chassis measures 36-cm wide × 18-cm high. Fins of aluminum (k = 205 W/m-K) that are 7.5-cm high, 2.286-mm thick, and 18-cm long can be used to enhance the heat transfer capability of the plate. The surrounding fluid is at 100 kPa and 22◦ C and the chassis temperature is 95◦ C. Determine the number of fins to be employed and the heat dissipation. 24.59: Eight vertical plates are placed x cm apart in the air at 100 kPa and 22◦ C. The plates are 16 cm × 16 cm and are held at 67◦ C. Determine the total heat flow from the plates when (a) x = 2 cm, (b) x = 1.5 cm, and (c) x = 1 cm. 24.60: The dissipation from a transformer is improved by equipping it with six vertical fins of aluminum (k = 210 W/m-K) that are each 8-cm high, 4-mm thick, and 48-cm long. The fin centerline-to-centerline spacing is 2 cm, and the surface temperature of the transformer is held at 77◦ C. The surrounding air pressure is at 100 kPa. Determine the air temperature if 160 W are to be dissipated.

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24.61: Determine the heat transfer coefficient in natural convection within a 15-cm high × 7.5-cm wide parallel plate channel with a spacing of 2 cm. The air is at 100 kPa and 300 K, and the channel walls are maintained at 350 K. Enclosures 24.62: A vertical enclosure consists of two plates that are 20-cm high at a plate spacing of 2 cm. The plate on the left is held at 85◦ C and the one on the right is held at 25◦ C. Determine the heat transfer coefficient for the gap. 24.63: Rework Problem 24.62 for a plate height of 35 cm at a gap spacing of 1.4 cm. All other conditions remain the same. 24.64: Rework Problem 24.62 for a tilted enclosure at 53.13◦ . All other conditions remain the same. 24.65: Two concentric cylinders form a horizontal annular region with di = 2 cm and do = 4 cm. The temperatures of the inner and outer walls are 57◦ C and 17◦ C, respectively, and air at 100 kPa fills the annular region. Determine the heat flow for unit length. 24.66: Rework Problem 24.65 for an annular region with an unknown outer diameter but with heat flow across the gap of 16.75 W/m. All other conditions remain the same. 24.67: Rework Problem 24.65 for a pair of concentric spheres. All other conditions remain the same. 24.68: Rework Problem 24.62 for plate spacings of (a) 2.5 cm, (b) 4 cm, and (c) 5 cm. All other conditions remain the same. 24.69: Rework Problem 24.65 for outer cylinder diameters of (a) 4.5 cm, (b) 5 cm, and (c) 6.25 cm. All other conditions remain the same. 24.70: Rework Problem 24.69 for outer sphere diameters of (a) 4.5 cm, (b) 5 cm, and (c) 6.25 cm. All other conditions remain the same.

25 Heat Exchangers

Chapter Objectives •

To describe the types of heat exchangers and to delineate the differences between the recuperator, direct contact heat exchanger, and the regenerator.

•

To provide the governing relationships for the unfinned heat exchanger.

To consider the various forms of the overall heat transfer coefficient. • To develop the concept of the logarithmic mean temperature difference. •

•

To give the details of the logarithmic mean temperature difference correction factor method of heat exchanger design.

•

To give the details of the effectiveness Ntu method of heat exchanger design.

•

To provide the governing relationships for finned heat exchangers.

25.1

Introduction

A heat exchanger is a device that transfers heat from one fluid to another or between a fluid and the environment. In the open type, or direct contact type of heat exchanger, there is no intervening surface between the fluids. In the closed type or indirect contact heat exchangers, our definition pertains to a device that is employed in the transfer of heat between two fluids or a surface and a fluid. Shah (1981) and Mayinger (1988) have categorized heat exchangers according to the following properties: • • • •

Transfer processes Number of fluids Construction Heat transfer mechanisms

• • • •

Surface compactness Flow arrangement Number of fluid passes Type of surface

Recuperators (or closed-type heat exchangers) are heat exchangers in which heat transfer occurs between two fluid streams at different temperatures that are separated by a thin solid wall (a parting sheet or tube wall). Heat is transferred by convection from the hotter fluid to the wall surface and by convection from the wall surface to the cooler fluid. Figure 25.1 shows four examples of the recuperator that are surface-type exchangers. The recuperator is the only type of heat exchanger that we will consider in detail. Regenerators are heat exchangers in which a hot fluid and a cold fluid alternately flow through the same surface at prescribed time intervals. The surface of the regenerator receives 805

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Fluid-1

Fluid-1

Fluid-2

Fluid-2

(a) Fluid-1

Fluid-2

(b)

Fluid-1

Fluid-2

Fluid-1

(c)

Fluid-1

Fluid-2

Fluid-1

(d)

FIGURE 25.1 Examples of closed-type heat exchangers or recuperators. (a) The double-pipe counterflow heat exchanger, (b) the compact heat exchanger, (c) the shell and tube heat exchanger with one shell pass and two tube passes, and (d) the shell and tube heat exchanger with one shell pass and four tube passes.

Heat Exchangers

807

heat by convection from the hot fluid and then releases heat by convection to the cold fluid. The process is transient, that is, the temperature of the surface (and the fluids themselves) vary with time during the heating and cooling of the common surface. The regenerator is also a surface-type heat exchanger. In direct contact (or open type) heat exchangers, heat is transferred by partial or complete mixing of the hot and cold fluid streams. Hot and cold fluids that enter this type of exchanger separately leave together as a single mixed stream.

25.2

Governing Relationships

25.2.1 The Rate Equation We will assume that there are two process streams in a heat exchanger, a hot stream flowing with a capacity rate C H = m ˙ H c p, H and a cooler (or cold stream) flowing with a capacity rate CC = m ˙ C c p,C . Then, conservation of energy demands that the heat transferred between the streams be described by the enthalpy balance ˙ = C H (T1 − T2 ) = CC (t2 − t1 ) Q

(25.1)

where the subscripts 1 and 2 refer to the inlet and outlet of the exchanger and where the T’s and t’s are employed to indicate hot and cold fluid temperatures, respectively. Equation 25.1 represents an ideal that must hold in the absence of losses and, while it describes the heat that will be transferred (the duty of the heat exchanger) for the case of prescribed flow and temperature conditions, it does not provide an indication of the size of the heat exchanger necessary to perform this duty. We find that the size of the exchanger is determined using what is called the rate equation ˙ = U Sm = UH SH m = UC SC m Q

(25.2)

where SH and SC are the surface areas on the hot and cold sides of the exchanger, UH and UC are the overall heat transfer coefficients referred to the hot and cold sides of the exchanger and m is a driving temperature difference. The entire heat exchange process can be represented by ˙ = UH SH m = UC SC m = C H (T1 − T2 ) = CC (t2 − t1 ) Q

(25.3)

which is merely a combination of Equations 25.1 and 25.2. 25.2.2 The Exchanger Surface Area Consider the unfinned tube of length L shown in Figure 25.2 and observe that because of the tube wall thickness, w , the inner diameter will be smaller than the outer diameter and the surface areas will be different Si = di L

(25.4a)

So = do L

(25.4b)

and

where do = di + 2w .

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Introduction to Thermal and Fluid Engineering

L

δw

FIGURE 25.2 A heat exchanger tube of length, L .

25.2.3 The Overall Heat Transfer Coefficient In a heat exchanger involving hot and cold streams, heat must flow, in turn, from the hot fluid to the cold fluid through as many as five thermal resistances: •

The hot-side convective layer resistance RH =

•

1 h H SH

(K/W)

(25.5)

The hot-side fouling resistance due to an accumulation of foreign (and undesirable) material on the hot fluid exchanger surface. Rd, H =

1 h d, H SH

(K/W)

(25.6)

where the subscript d is intended to infer “dirt.” Fouling is discussed in Section 25.2.5. •

The resistance of the exchanger material that has a finite thermal conductivity and may take on a value that is a function of the type of exchanger

Rm =

⎧ ⎪ ⎪ ⎨

m km Sm

(K/W)

plane walls (25.7)

⎪ ⎪ ⎩ ln(do /di ) 2km Lnt

(K/W)

circular tubes

where m is the thickness of the wall, Sm is the surface area of the wall, and nt is the number of tubes. •

The cold-side fouling resistance Rd,C =

•

1 h d,C SC

(K/W)

(25.8)

The cold-side convective layer resistance RC =

1 h C SC

(K/W)

(25.9)

Heat Exchangers

809

The resistances listed in Equations 25.5 through 25.9 are in “series” and the total resistance can be represented by RT = 1/U S 1 1 1 1 1 = + + Rm + + US h H SH h d, H SH h d,C SC h C SC

(25.10)

where, for the moment, U and S on the left side of Equation 25.10 are not assigned subscripts. We see that Equation 25.10 is perfectly general and may be put into specific terms depending on the selection of the reference surface, whether or not fouling is present and whether or not the conduction resistance need be considered. If Equation 25.10 is solved for U, the result is U=

1 S S S S + + SRm + + h H SH h d, H SH h d,C SC h C SC

(25.11)

and if the thickness of the tube material is small or its thermal conductivity is high, the conduction resistance becomes negligible and U=

1 S S S S + + + h H SH h d, H SH h d,C SC h C SC

(25.12)

Several forms of Equation 25.12 are •

For a hot-side reference with fouling on both sides (S = SH ) UH =

•

1 1 SC 1 SC 1 1 + + + h H SH h d, H SH h d,C hC

(25.14)

For a hot-side reference without fouling (S = SH ) UH =

•

(25.13)

For a cold-side reference with fouling on both sides (S = SC ) UC =

•

1 1 1 1 SH 1 SH + + + hH h d, H h d,C SC h C SC

1 1 1 SH + hH h C SC

(25.15)

For a cold-side reference without fouling (S = SC ) UC =

1 1 SC 1 + h H SH hC

(25.16)

A listing of some approximate overall heat transfer coefficients for various services is provided in Table 25.1.

Example 25.1 Hot water flows through a 3-in-schedule 80 pipe with an inside heat transfer coefficient of 1800 W/m2 -K. It is cooled by air with a natural convection heat transfer

810

Introduction to Thermal and Fluid Engineering TABLE 25.1

Approximate Overall Heat Transfer Coefficients for Some Process Services Application

U, W/m2 -K

Water to water Water to air Water to lube oil Water to organic vapor Water to heavy fuel oil Heavy organics to heavy organics Light organics to light organics Gas to gas Condensing steam to water Condensing steam to air Refrigerator evaporators

750–2500 7.5–50 100–400 250–1000 50–200 40–200 180–450 8–30 1000–6000 50–200 250–1000

coefficient of 8 W/m2 -K, and there is no fouling. Determine the heat lost per unit of length per degree Celsius temperature difference driving force based on (a) the outside surface and (b) the inside surface.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. We first note that the hot fluid (the water) flows through the pipe with an inside diameter, di . This leaves the cold fluid (the air) on the outside of the pipe with diameter, do . The ratio of the cold side to the hot-side surface areas will be SC do L do = = SH di L di Table A.20 in Appendix A gives for the 3-in-schedule 80 pipe do = 8.890 cm

and

di = 7.366 cm

so that do SC 8.890 cm = = = 1.207 di SH 7.366 cm and di 1 = = 0.829 do 1.207 (a) Because the cold fluid (the air) is on the outside of the pipe, use Equation 25.16 with SC /SH = do /di = 1.207 UC =

1 1 = 1 SC 1 1.207 1 + + 2 h H SH hC 1800 W/m -K 8 W/m2 -K

Heat Exchangers

811

or UC =

1 2

(6.705 × 10−4 W/m -K) −1 + (0.125 W/m2 -K) −1

= 7.96 W/m2 -K

Then, for a surface area of SC = do L = (0.0889 m)(1 m) = 0.2793 m2 we have ˙ Q = UC SC = (7.96 W/m2 -K)(0.2793 m2 ) = 2.22 W/K ⇐ m (b) With the hot fluid (the water) on the inside of the pipe, we use Equation 25.15 with SH /SC = di /do = 0.829 UH =

1 1 = 1 1 SH 1 0.829 + + 2 hH h C SC 1800 W/m -K 8 W/m2 -K

or UH =

1 (5.556 ×

10−4

2

W/m -K) −1 + (0.104 W/m2 -K) −1

= 9.60 W/m2 -K

Then, for a surface area of SH = di L = (0.0737 m)(1.00 m) = 0.2314 m2 we have ˙ Q = UH SH = (9.60 W/m2 -K)(0.2314 m2 ) = 2.22 W/K ⇐ m 25.2.4 The Logarithmic Mean Temperature Difference We now focus our attention on the value of m in Equation 25.2 and in Figure 25.3, which shows the temperature length profile for the two fluids in a parallel or cocurrent flow heat exchanger. It appears that various alternatives are available to us. We could use m = T1 −t1 ; however, this temperature difference is the maximum difference available, its use would ˙ On the other hand, use of be optimistic and would lead to erroneously high values of Q. m = T2 − t2 , which is the minimum temperature difference, would be pessimistic and lead ˙ that are too low. Of course, we could take an arithmetic average, but this to values of Q would need verification. In working with Figure 25.3 we will assume that •

The overall coefficient of heat transfer, U, is constant throughout the entire heat exchanger.

•

There is no heat lost to the surroundings.

•

There is no longitudinal heat conduction in the direction of the flowing fluids.

•

There is no phase change (or constant temperature zone) in either fluid.

812

Introduction to Thermal and Fluid Engineering Trt T1 dT T–t

T2 t2 dt

t1

S 1

2

dS

FIGURE 25.3 The hot and cold temperature profiles for the fluids in a parallel flow heat exchanger of length, L . The inlet and outlet end of the exchanger are designated as points 1 and 2, respectively. The profiles are plotted as a function of the surface area, S, and the differential element of surface area may noted. These are used to develop the LMTD for this type of exchanger.

•

The specific heats, flow rates (and hence, the capacity rates) for both fluids are constant.

•

The fluid temperatures at any cross section are characterized by the two temperatures, T and t.

˙ can be related to the For the differential element of surface area, dS, the heat flow, d Q, enthalpy change of each fluid ˙ = −C H dT = CC dt dQ

(25.17)

and the rate equation applied to the differential element shown in Figure 25.3 provides ˙ = U(T − t)dS dQ

(25.18)

From Equation 25.17 we have dT = −

˙ dQ CH

and

and a subtraction shows that ˙ dT − dt = d(T − t) = −d Q

dt = 

˙ dQ CC

1 1 + CH CC

 (25.19)

Equations 25.18 and 25.19 may then be combined to obtain   d(T − t) 1 1 dS = −U + T −t CH CC and an integration between point 1 where T1 = T1 − t1 and point 2 where T2 = T2 − t2 will yield     T2 − t2 1 1 ln = −U S (25.20) + T1 − t1 CH CC

Heat Exchangers

813

Now with CH = −

˙ Q T1 − T2

Equation 25.20 may be displayed as

˙ Q t2 − t1

CC =

and

⎡

⎤

⎢ t2 ) − (T1 − t1 ) ⎥ ⎥ ˙ = U S ⎢ (T2 −   Q ⎣ ⎦ T2 − t2 ln T1 − t1

(25.21)

Then a comparison of Equations 25.2 and 25.21 gives m for the parallel flow heat exchanger in Figure 25.3. Because of the natural logarithm in the denominator, m is referred to as the logarithmic mean temperature difference and designated merely as LMTD. m = LMTD =

(T2 − t2 ) − (T1 − t1 )   T2 − t2 ln T1 − t1

(25.22)

Similar derivations for the four basic simple arrangements indicated in Figure 25.4 will provide the logarithmic mean temperature difference that can be written as m = LMTD =

t1

T2

T1 − T2 T2 − T1 = T1 T2 ln ln T2 T1 T1

t2

t1

T

T

T1

T1

t2

(25.23)

T1

T2

t2

T2 t2

T2 t1

t1 L

L

(a)

(b)

Ts

T1 t2

T2

ts

t1 L (c)

L (d)

FIGURE 25.4 Temperature profiles for the four basic heat exchanger arrangements in which the LMTD may be employed. The counterflow heat exchanger, (b) the parallel or cocurrent flow heat exchanger, (c) the exchanger with a constant temperature source fluid and a rising temperature receiver fluid, and (d) the exchanger with a constant temperature receiver fluid and a falling temperature source fluid.

814 •

Introduction to Thermal and Fluid Engineering For the counterflow exchanger where the fluids flow in opposite directions through the exchanger (Figure 25.4a) LMTD =

•

(T1 − t1 ) − (T2 − t2 ) T1 − t1 ln T2 − t2

(25.22)

For an exchanger that has a constant temperature source, Ts = T1 = T2 , and a rising temperature receiver (Figure 25.4c) LMTD =

•

(25.24)

For the cocurrent or parallel flow exchanger where the fluids flow in the same direction through the exchanger (Figure 25.4b) LMTD =

•

(T1 − t2 ) − (T2 − t1 ) T1 − t2 ln T2 − t1

t2 − t1 Ts − t1 ln Ts − t2

(25.25)

For an exchanger that has a constant temperature receiver, ts = t1 = t2 , and a falling temperature source (Figure 25.4d) LMTD =

T1 − T2 T1 − ts ln T2 − ts

(25.26)

The logarithmic mean temperature difference must not be employed for arrangements other than those shown in Figure 25.4. The procedure for the case of cross flow and shell and tube exchangers is given in the next section.

Example 25.2 A process stream is cooled from 320◦ C to 160◦ C by another stream that is heated from 80◦ C to 140◦ C (Figure 25.5). Determine the LMTD for (a) counterflow and (b) cocurrent or parallel flow.

T1 T2 T2 t2 t1 T1

t1

t2 (a) FIGURE 25.5 Temperature profiles for Example 25.2.

(b)

Heat Exchangers

815

Solution Assumptions and Specifications (1) None. Here, we are to find the LMTD for two flow arrangements with given temperatures at inlet and outlet for both streams. We have T1 = 320◦ C

and

T2 = 160◦ C

t1 = 80◦ C

and

t2 = 140◦ C

and

(a) For counterflow, use Equation 25.24 LMTD =

=

=

(T1 − t2 ) − (T2 − t1 ) T1 − T2 ln T2 − t1 (320◦ C − 140◦ C) − (160◦ C − 80◦ C) 320◦ C − 140◦ C ln 160◦ C − 80◦ C 180◦ C − 80◦ C 100◦ C = = 123.3◦ C ◦ 180 C ln 2.25 ln 80◦ C

(b) For cocurrent or parallel flow, use Equation 25.22 LMTD =

=

=

(T1 − t1 ) − (T2 − t2 ) T1 − t1 ln T2 − t2

(320◦ C − 80◦ C) − (160◦ C − 140◦ C) 320◦ C − 80◦ C ln 160◦ C − 140◦ C 240◦ C − 20◦ C 220◦ C = = 88.5◦ C 240◦ C ln 12 ln 20◦ C

We can see that the parallel flow LMTD is much lower than the counterflow LMTD. In fact, it can be proven that the best possible LMTD for parallel flow reaches the limit of the counterflow LMTD. You might ask why, under these circumstances, is parallel flow ever used? The answer lies in the fact that cramped space in a proposed application may not permit the piping arrangement for counterflow and it may be desirable, in attempting to heat a highly viscous fluid, to provide a high-temperature difference at the front end to get the fluid flowing. 25.2.5 Fouling 25.2.5.1 Fouling Mechanisms Somerscales and Knudsen (1981) have identified six categories of fouling: 1. Particulate fouling: The accumulation of solid particles suspended in the process stream on the heat transfer surfaces.

816

Introduction to Thermal and Fluid Engineering

2. Precipitation fouling: The precipitation of dissolved substances carried in the process stream upon the heat transfer surfaces. Scaling occurs when precipitation occurs on heated rather than cooled surfaces. 3. Chemical reaction fouling: In certain cases, deposits on the heat transfer surfaces which are not, in themselves, reactants are formed by chemical reactions. 4. Corrosion fouling: In this type of fouling, the heat transfer surface reacts, at certain pH levels, to produce products that adhere to the heat transfer surfaces and, in turn, this may promote the attachment of additional fouling materials. 5. Biological fouling: Materials such as algae, bacteria, molds, seaweed, and barnacles carried in the process stream cause biological fouling of the heat transfer surfaces. Marine power plant condensers are notorious for their vulnerability to biological fouling. 6. Freezing fouling: Sometimes a liquid, or some of its higher melting point components, will deposit upon a subcooled heat transfer surface. 25.2.5.2 Fouling Factors When a heat exchanger is placed in service, the heat transfer surfaces are, presumably, clean. With time, in some services in the power and process industries, the apparatus may undergo a decline in its ability to transfer heat. This is due to the accumulation of heat insulating substances on either or both of the heat transfer surfaces. The Tubular Exchanger Manufacturers Association (TEMA) published a table of fouling factors to assist the designer. Fouling resistances were tabulated and these were to be added to the film resistances (1/Si h i and 1/ h o So ) of specific process streams. In this way, the operating period of all heat exchangers would be similar and ensure some desired period of continuous operation. The tables of fouling factors were intended as a crude guide toward the equalization of cumulative fouling in all fouling streams in the assembly. Extracts from the table of fouling factors given by Chenoweth (1990), but in terms of 1/ h d , are provided in Table 25.2.

Example 25.3 A heat exchanger employs 2-in OD × 12-BWG (Birmingham Wire Gage) tubes that have outer and inner diameters of 5.080 and 4.526 cm, respectively. The tubes are 4-m long and carry a hot process stream on the inside with h H = 800 W/m2 -K and a cold fluid stream on the outside with h C = 500 W/m2 -K. Determine the overall heat transfer coefficient referred to the cold side when (a) the exchanger is clean and (b) when the exchanger is fouled to the extent that a fouling resistance of 0.0008 (W/m2 -K)−1 is present on the hot side.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. First, we need to determine SH and SC . The hot fluid is in the inside of the tubes SH = di L and hence, SC = do L

Heat Exchangers

817 TABLE 25.2

Some Fouling Resistances of Various Gas, Vapor, and Liquid Streams (Adapted from Chenoweth, 1990) Fouling Resistance 1 × 104 , m2 K/W hd

Fluid Cooling tower water Artificial spray pond water Brackish water Treated boiler feedwater Seawater No. 2 Fuel oil Engine lube oil Refrigerants (liquid) Gasolene Heavy fuel oil Natural gas Compressed air Refrigerant vapor (oil bearing) Steam (non-oil bearing)

1.75–3.50 1.75–3.50 3.50–5.25 0.90 1.75–3.50 3.50 1.75 1.75 3.50 5.25–12.50 1.75–3.50 1.75 3.50 9.00

Thus, SC do L do = = SH di L di With the given values that can be verified in Table A.21 do = 5.080 cm

di = 4.526 cm

and

we have do SC 5.080 cm = = = 1.122 di SH 4.526 cm (a) Because the cold fluid is on the outside of the pipe, use Equation 25.16 with SC /SH = do /di = 1.207 UC =

=

1 1 = 1 SC 1 1.122 1 + + 2 h H SH hC 800 W/m -K 500 W/m2 -K 1 2

(1.403 × 10−3 W/m -K) −1 + (2 × 103 W/m2 -K) −1

= 293.9 W/m2 -K ⇐ (b) When a fouling factor of 1/ h d, H = 0.0008 (W/m2 K) 1 is employed on the hot side, we use Equation 25.14 UC =

1 1 SC 1 SC 1 1 + + + h H SH h d, H SH h d,C hC

818

Introduction to Thermal and Fluid Engineering

and with 1/ h d,C = 0, we have UC =

1 1.122 2

800 W/m -K =

+ (0.0008 m2 -K/W) (1.122) +

1 500 W/m2 -K

1 [1.403 ×

10−3

+ 8.976 × 10−4 + 2 × 10−3 ] W/m2 -K

= 232.6 W/m2 -K ⇐

25.3

Heat Exchanger Analysis Methods

25.3.1 The Logarithmic Mean Temperature Difference Correction Factor Method The logarithmic mean temperature difference developed in Section 25.2.4 is not applicable to shell and tube or cross-flow heat exchangers. In these configurations, the temperature parameter m in Equations 25.2 and 25.3 is the true or effective mean temperature difference and is related to the counterflow logarithmic mean temperature difference LMTDc =

(T1 − t2 ) − (T2 − t1 ) T1 − T2 ln T2 − t1

(25.24)

via m = F (LMTDc )

(25.27)

where F is a factor that is referred to as the logarithmic mean temperature difference correction factor. It is a function of two parameters, P and R where P=

t2 − t1 T1 − t1

(25.28a)

is defined as the cold-side effectiveness and R=

T1 − T2 CC = t2 − t1 CH

(25.28b)

is a capacity rate ratio. The evaluation of the the logarithmic mean temperature difference correction factor for the multitude of multipass and cross-flow heat exchanger arrangements dates from the early 1930s: Nagle (1933), Underwood (1934), Fischer (1938), and Bowman et al. (1940). The correction factors are available in chart form such as those shown in Figures 25.6 through 25.9.

Example 25.4 In a water-to-water shell and tube heat exchanger, hot water is on the shell

side and flows at 2.655 kg/s at an inlet temperature of 95◦ C. Cold water flows within the tubes at a rate of 4.00 kg/s between temperatures of 40◦ C and 60◦ C. The overall heat transfer coefficient referred to the hot side of the exchanger is 1250 W/m2 -K. The exchanger may be considered unfouled with tubes that are to be limited to a maximum length of 2.25 m. If

Heat Exchangers

819

1.0

0.8

0.8

0.9

0.1

0.3

0.7

0.4

0.6

0.5

0.5

0.6

0.7 0.8 0.9 1.0

1.2

0.4

1.4

0.3

1.6 1.8 2.0

2.5

0.2

3.0

0.1

4.0

0

6.0

0.5

8.0 10.0 15.0

0.6

0.2

0.7 R = 20.0

Correction Factor, F

0.9

1.0

Temperature Efficiency, P P=

t2 – t1 T1 – t1

R=

T1 – T2 t2 – t1

FIGURE 25.6 The logarithmic mean temperature difference correction factor F for a shell and tube heat exchanger with one shell pass and two tube passes. (From Bowman et al., 1940.)

the tubes are 3/4-in OD × 18 BWG and the velocity within the tubes is to be maintained as close as possible to 0.425 m/s, determine the number of tube passes and the number and length of the tubes.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. (5) The flow of the water in the tubes is fully developed. (6) Fluid specific heats and the overall heat transfer coefficient are constant. (7) Axial conduction along the exchanger tubing is negligible. We first assume a single pass on the tube side and we will check this assumption against the maximum tube length of 2.25 m. The hot-side outlet temperature is obtained from a rearrangement of Equation 25.1 T2 = T1 −

m ˙ C c C (t2 − t1 ) m ˙ HcH

820

Introduction to Thermal and Fluid Engineering 1.0

0.1 0.3

0.9

0.2

1.6

Correction Factor, F

0.4 0.5 0.6 0.7 0.9

0.8

1.0

1.2

1.4

1.8 2.0

2.5

3.0

4.0

6.0

0.7

8.0 10.0 15.0

R = 20.0

0.8

0.6

0.5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Temperature Efficiency, P

P=

t2 – t1 T1 – t1

R=

T1 – T2 t2 – t1

FIGURE 25.7 The logarithmic mean temperature difference correction factor F for a shell and tube heat exchanger with two shell passes and four tube passes. (From Bowman et al., 1940.)

Here, we retain the subscripts because of the variation of specific heat with the temperature of the water. The average cold water temperature is

tav =

t1 + t2 40◦ C + 60◦ C = = 50◦ C 2 2

and Table A.14 gives c C = 4.178 kJ/kg-K When we assume T2 = 65◦ C (which will also require verification) we find that the average hot water temperature will be

Tav =

T1 + T2 95◦ C + 65◦ C = = 80◦ C 2 2

and Table A.14 gives c H = 4.197 kJ/kg K

Heat Exchangers

821

1.0

R= 0.2

0.4

0.8 0.6 0.8

1.0

1.5

0.7

2.0

Correction Factor, F

0.9

3.0

4.0

0.6

0.5

0

0.1

0.2

0.3

0.4

0.6

0.7

0.8

0.9

1.0

Temperature Efficiency, P

T1 P= t1

0.5

t2 – t1 T1 – t1

R=

T1 – T2 t2 – t1

t2

T2 FIGURE 25.8 The logarithmic mean temperature difference correction factor F for a cross-flow heat exchanger with both fluids unmixed. (From Bowman et al., 1940.)

Thus, T2 = T1 −

m ˙ C c C (t2 − t1 ) m ˙ HcH

= 95◦ C −

(4 kg/s)(4.178 kJ/kg-K)(60◦ C − 40◦ C) (2.655 kg/s)(4.197 kJ/kg-K)

= 95◦ C − 30◦ C = 65◦ C This is indeed the assumed value and permits the calculation procedure to continue. The heat duty may also be obtained from Equation 25.1. Working with the cold side gives ˙ =m Q ˙ C c C (t2 − t1 ) = (4 kg/s)(4.178 kJ/kg-K)(60◦ C − 40◦ C) = 334.2 KW

822

Introduction to Thermal and Fluid Engineering 1.0

0.8

0.4

0.6

0.8

1.0

1.5

2.0

3.0

R = 0.2

0.7 4.0

Correction Factor, F

0.9

0.6

0.5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Temperature Efficiency, P T1 P=

t1

t2 – t1 T1 – t1

R=

T1 – T2 t2 – t1

t2

T2 FIGURE 25.9 The logarithmic mean temperature difference correction factor F for a cross-flow heat exchanger with one fluid mixed and the other fluid unmixed. (From Bowman et al., 1940.)

For the one shell pass-one tube pass exchanger, we have true counterflow and we employ Equation 25.24 for the LMTD LMTDc =

(T1 − t2 ) − (T2 − t1 ) (95◦ C − 60◦ C) − (65◦ C − 40◦ C) = T1 − T2 95◦ C − 60◦ C ln ln ◦ T2 − t1 65 C − 40◦ C

or LMTDc =

35◦ C − 25◦ C 10 = = 29.72◦ C 35◦ C ln 1.40 ln ◦ 25 C

Because of the cold-side velocity requirement, we need to determine the total inside or cold-side cross-sectional area. At tav = 50◦ C, Table A.14 gives C = 988.1 kg/m3 and then we have AC =

m ˙C 4.00 kg/s = = 9.525 × 10−3 m2 ˆ (988.1kg/m3 )(0.425 m/s) C V C

Heat Exchangers

823

Table A.21 shows that for a 3/4-in × 18-BWG tube do = 1.905 cm and

di = 1.656 cm

and it may be shown that Ai = 2.154 × 10−4 m2

So = 0.05202 m2 / m

and

where So is the outside surface of the tube in m2 / m. To maintain a “come as close as possible to” velocity of 0.425 m/s, we have nt =

AC 9.525 × 10−3 m2 = = 44.22 Ai 2.154 × 10−4 m2

(say 44 tubes)

and with the tubes at their maximum length of 2.25 m, we have SH = nt So L = (44 tubes)(0.05202 m2 / m)(2.25 m) = 5.150 m2 Equation 25.2 with m = LMTD now gives ˙ = UH SH LMTD = (1250 W/m2 -K)(5.150 m2 )(29.72◦ C) = 191.34 kW Q This falls far short of the required duty of 334.2 Kwa. Because the single-pass heat exchanger does not perform the required duty, we must go to two passes still maintaining 44 tubes per pass in order to satisfy the water velocity requirement. But when we go to the two-pass exchanger, we must make a correction to the LMTD in accordance with Equation 25.27. This correction will require calculation of the values of P and R in accordance with Equation 25.28 P=

t2 − t1 60◦ C − 40◦ C 20◦ C = ◦ = ◦ = 0.364 ◦ T1 − t1 95 C − 40 C 55 C

R=

T1 − T2 95◦ C − 65◦ C 30◦ C = ◦ = = 1.50 t2 − t1 60 C − 40◦ C 20◦ C

and

Figure 25.6 then provides F = 0.88 ˙ = 334.2 Kwa and nt = 2(44) = 88 tubes and with Q SH =

˙ Q 334.2 kW = = 10.22 m2 UH F (LMTDc ) (1250 W/m2 -K)(0.88)(29.72◦ C)

The tube length will be L=

SH 10.22 m2 = = 2.23 m < 2.25 m nt do (88 tubes)(0.05202 m2 / m)

Thus, there will be 2 passes

⇐

88 tubes ⇐ 2.23 m long ⇐

824

Introduction to Thermal and Fluid Engineering

25.3.2 The Effectiveness Ntu Method 25.3.2.1 Dimensionless Parameters In the logarithmic mean temperature difference correction factor method, the parameter P, requires three temperatures for its computation. The inlet temperature of both the hot and cold streams is usually a given; when the cold-side outlet temperature is not known, however, a trial-and-error method is required to evaluate P. The trial-and-error procedure may be avoided by using the -Ntu method and, because of its suitability for computer aided design, the -Ntu method is gaining in popularity. Kays and London (1984) have shown that the heat exchanger transfer equations may be written in a dimensionless form that results in three dimensionless groups: 1. The capacity rate ratio C∗ =

Cmin Cmax

(25.29)

Notice that this differs from the capacity rate ratio, R, used in the determination of the logarithmic mean temperature difference correction factor. Here, the capacity ratio, C ∗ , is always less than or equal to unity. It may take on the value C ∗ = 0 for the cases of the constant temperature source and rising temperature receiver fluid and the falling temperature source and constant temperature receiver fluid. 2. The exchanger heat transfer effectiveness =

˙ Q ˙ max Q

( < 1)

(25.30)

This is the ratio of the actual heat transferred to the maximum heat that could be transferred if the exchanger were a counterflow exchanger. We note that the limit of the effectiveness for an exchanger of infinite size would be unity. 3. The number of transfer units US 1 Ntu ≡ = Cmin Cmin

 U dS

(25.31)

S

The number of transfer units is a measure of the size of the exchanger. The actual heat transfer is given by the enthalpy balance of Equation 25.1. Observe that if C H > CC

then

(T1 − T2 ) < (t2 − t1 )

CC > C H

then

(t2 − t1 ) < (T1 − T2 )

and if

the fluid that “might” experience the maximum temperature change, T1 −t1 , is the fluid that has the minimum capacity rate. Thus, the maximum possible heat transfer can be expressed as ˙ max = CC (T1 − t1 ) Q

(CC < C H )

(25.32a)

˙ max = C H (T1 − t1 ) Q

(C H < CC )

(25.32b)

or

Heat Exchangers

825

and either of these can be obtained with the counterflow exchanger. Therefore, the exchanger effectiveness can be written as

=

˙ Q C H (T1 − T2 ) CC (t2 − t1 ) = = ˙ max Cmin (T1 − t1 ) Cmin (T1 − t1 ) Q

(25.33)

Observe that the value of  will range between zero and unity and that for a given  and ˙ max , the actual heat transfer in the exchanger will be Q ˙ = Cmin (T1 − t1 ) Q

(25.34)

Because  = f (C ∗ , Ntu , flow arrangement) each exchanger arrangement has its own effectiveness relationship. Figure 25.10 provides graphs for four such arrangements. 25.3.2.2 Specific Effectiveness -Ntu Relationships Specific  Ntu relationships along with their limiting values for seven flow arrangements, summarized from the work of Kays and London (1984) and Kakac et al. (1987) now follow: 1. For counterflow ∗

=

1 − e −Ntu (1−C ) 1 − C ∗ e −Ntu (1−C ∗ ) =

Ntu Ntu + 1

(C ∗ < 1)

(25.35a)

(C ∗ = 1)

(25.35b)

1 1 − C ∗ ln 1 − C∗ 1−

(25.36)

and Ntu = 2. For cocurrent or parallel flow =

1 − e −Ntu (1+C 1 + C∗

∗

)

(25.37)

and Ntu =

1 1 ln 1 + C ∗ 1 − (1 + C ∗ )

(25.38)

3. For single-pass cross flow with both fluids unmixed 0.22

/C ∗

(Ntu ) 0.78

−1

 = 1 − e (Ntu ) where  = e −C

∗

Equation 25.39 cannot be rearranged to yield Ntu = f (, C ∗ ).

(25.39)

826

Introduction to Thermal and Fluid Engineering Shell Fluid (mcp)s = Cs

Shell Fluid (mcp)s = Cs Two Shells Tube Fluid (mcp)t = Ct One Shell Passes 2, 4, 6, etc., Tube Passes Cmin/Cmax = 0

100

0.50 0.75 1.00

60

0.25 0.50 0.75 1.00

80 Effectiveness, ε, %

Effectiveness, ε, %

100

0.25

80

Tube Fluid (mcp)t = Ct Two Shell Passes 4, 8, 12, etc., Tube Passes Cmin/Cmax = 0

40 20

60

40 20

0 0

1

2

3

4

0 0

5

1

2

3

4

5

Number of Transfer Units, Ntu,max = US/Cmin

Number of Transfer Units, Ntu,max = US/Cmin

(a)

(b)

Cross-flow Exchanger with Fluids Unmixed (mcp)c

Cross-flow Exchanger with One Fluid Mixed

Cold Fluid Mixed Fluid

(mcp)h Hot Fluid 100

60

40

20

0

0.25 4 0.5 2 0.75 1.33

80

Effectiveness, ε, %

Effectiveness, ε, %

Cmixed = 0, ∞ Cunmixed

0.25 0.50 0.75 1.00

80

Unmixed Fluid

100

Cmin/Cmax = 0

60 Cmixed =1 Cunmixed 40

20 0 0

1

2

3

4

5

Number of Transfer Units, Ntu,max = AU/Cmin (c)

0

1

2

3

4

5

Number of Transfer Units, Ntu,max = US/Cmin (d)

FIGURE 25.10 The effectiveness  as a function of the number of transfer units, Ntu , for (a) the shell and tube exchanger with one shell pass and two tube passes, (b) the shell and tube exchanger with two shell passes and four tube passes, (c) the cross-flow heat exchanger with both fluids unmixed, and (d) the cross-flow heat exchanger with one fluid mixed and one fluid unmixed.

Heat Exchangers

827

4. For single-pass cross flow with both fluids mixed 

1 C∗ 1 = + − 1 − e −Ntu 1 − e −C ∗ Ntu Ntu

−1 (25.40)

Equation 25.40 cannot be rearranged to yield Ntu = f (, C ∗ ). 5. For single-pass cross flow with Cmax mixed and Cmin unmixed =

1 (1 − ) C∗

(25.41)

where  = e −C and

∗

(1−e −Ntu )

  1 Ntu = − ln 1 + ∗ ln((1 − C ∗ ) C

(25.42)

6. For single-pass cross flow with Cmin mixed and Cmax unmixed  = 1 − e

(25.43)

where  1  ∗ 1 − e −C Ntu C∗

=− and Ntu = −

1 ln[C ∗ ln(1 − ) + 1] C∗

(25.44)

7. For the shell and tube heat exchanger with one shell pass and 2, 4, 6, . . ., tube passes   −Ntu (1+C ∗2 ) 1/2 1 + e  = 1 = 2 1 + C ∗ + (1 + C ∗2 ) 1/2 (25.45) 1 − e −Ntu (1+C ∗2 )1/2 and Ntu =

2/ − (1 + C ∗ ) (1 + C ∗2 ) 1/2

(25.46)

8. n Shell passes and 2n, 4n, 6n, . . ., tube passes  =

(1 − 1 C ∗ 1 − 1

n

   −1 (1 − 1 C ∗ n −1 − C∗ 1 − 1

(25.47)

where 1 is based on the Ntu per shell pass (Ntu /n) and

Ntu,1

2( F − C ∗ ) − (1 + C ∗ ) F − 1 = (1 + C ∗2 ) 1/2

(25.48)

828

Introduction to Thermal and Fluid Engineering where  F =

C ∗ − 1 −1

1/n

with Ntu,1 taken as the Ntu per shell pass (Ntu = nNtu,1 ).

Example 25.5 Determine the surface area for a cross-flow heat exchanger that will cool

8 kg/s of methyl alcohol (c = 2.470 kJ/kg-K) from 96◦ C to 70◦ C using 6.75 kg/s of water available at 12◦ C. The water flows on the inside of 1-in OD × 16-BWG tubes. Consider that the methyl alcohol is mixed and the overall heat transfer coefficient based on the outside tube surface area is 600 W/m2 -K.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. (5) The methyl alcohol is specified as mixed and the cooling water is assumed to be unmixed. (6) Fluid specific heats and the overall heat transfer coefficient are constant. (7) Axial conduction along the heat exchanger tubing is negligible. We are given the specific heat of the methyl alcohol and assume that it is the value at the average temperature. Thus, C H = (8.00 kg/s)(2.470 kJ/kg-K) = 19.76 kW/K and we can find the duty of the exchanger using Equation 25.1 ˙ = C H (T1 − T2 ) = (19.76 kW/K)(90◦ C − 70◦ C) = 395.2 kW Q We need the specific heat of the water and to find it, we assume that the water will leave the exchanger at t2 = 26◦ C, which we will need to verify. Then with tav =

t1 + t2 12◦ C + 26◦ C = = 19◦ C 2 2

From Table A.14 we find c C = 4.191 kJ/kg K and CC = m ˙ C c C = (6.75 kg/s)(4.191 kJ/kg-K) = 28.29 kW/K We then use Equation 25.1 to verify t2 t2 = t1 +

˙ Q 395.2 kW = 12◦ C + m ˙ C cC (6.75 kg/s)(4.191 kJ/kg-K)

= 12◦ C + 14.00◦ C = 26◦ C

Heat Exchangers

829

Hence, we take t2 = 26◦ C and we find that CC is correct at 28.29 kW/K. The foregoing values of C H and CC show that C H = Cmin and CC = Cmax and via Equation 25.29 C∗ =

Cmin 19.76 kW/K = = 0.699 Cmax 28.29 kW/K

It is indeed fortunate that C H = Cmin because the reference value of the overall heat transfer coefficient corresponds to Cmin . This eliminates the hassle of converting the overall heat transfer coefficient to the cold side of the exchanger. With C H = Cmin , Equation 25.33 gives us the effectiveness =

T1 − T2 96◦ C − 70◦ C 26◦ C = ◦ = = 0.310 T1 − t1 96 C − 12◦ C 84◦ C

Here, C H = Cmin is mixed and the value of Ntu required to give  = 0.310 with C ∗ = 0.699 is determined from Equation 25.44 Ntu = − =−

1 ln[C ∗ ln(1 − ) + 1] C∗ 1 ln[(0.699) ln(1 − 0.310) + 1] 0.699

= −(1.4306) ln [(0.699)(ln 0.690) + 1] = −(1.4306) ln [−2694 + 1] = −(1.4306)(ln 0.7406) = −(1.4306)(−0.3003) = 0.4296 We then find from Equation 25.31 that SH =

Ntu C H (0.4296)(19.76 KW/K) = = 14.15 m2 ⇐ UH 0.600 kW/m2 -K

Our attention now turns to a design example in which the heat transfer coefficient on both the hot and cold sides of the exchanger must be evaluated.

25.4

Design Example 17

A five-pass tubular air preheater with tubes having a maximum length of 7.0 m is to be used to heat 12 kg/s of air from 0◦ C to 204◦ C. Flue gas, having the properties of air, flows in the inside of the tubes at a rate of 12.93 kg/s and enters the tubes at 320◦ C. The air heater is to contain a bank of 2-in × 12-BWG tubing that is 48 tubes wide and 14 tubes deep on transverse and in-line spacings of ST = SL = 7.5 cm. The five air passes and one flue gas pass will permit the heater to be designed on the basis of true counterflow. Fouling of the surfaces may be considered negligible. Determine the total length of the tubing necessary to yield the 204◦ C outlet air temperature.

830

Introduction to Thermal and Fluid Engineering

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. (5) Fluid specific heats are constant. (6) Axial conduction along the heat exchanger tubing is negligible. (7) The air heater may be treated as an exchanger in counterflow. The bulk temperature of the air is tb =

t1 + t2 0◦ C + 204◦ C = = 102◦ C 2 2

(375 K)

and at this value, Table A.13 reveals (with interpolation) that c p,C = 1.0115 kJ/kg-K The duty of the air heater is therefore ˙ =m Q ˙ C c p,C (t2 − t1 ) = (12 kg/s)(1.0115 kJ/kg-K)(204◦ C − 0◦ C) = 2.476 MW Assume that the temperature of the flue gas at the exit is T2 = 134◦ C. This would yield an average bulk temperature for the flue gas of Tb =

T1 + T2 320◦ C + 134◦ C = = 227◦ C 2 2

(500 K)

and we find from Table A.13 that c p,H = 1.0295 kJ/kg-K provides an assumed heat duty of ˙ =m Q ˙ h c p,H (T1 − T2 ) = (12.93 kg/s)(1.0295 kJ/kg-K)(320◦ C − 134◦ C) = 2.476 MW The actual duty determined from the cold air side must match the assumed duty determined from the hot flue gas side. This means that the temperature of the air leaving the heater is indeed 134◦ C and demonstrates, once again, the value of the heat balance.

Heat Exchangers

831

The thermal properties for both the flue gas and the air may be obtained from Table A.13: Flue Gas Hot Side 500 0.0404 1.0295 2.671 0.680 0.7048

Air Cold Side Tb /tb , K k, W/m-K c p , kJ/kg-K  × 105 , kg/m-s Pr , kg/m3

375 0.0318 1.0115 2.180 0.693 0.9403

and we estimate Pr at the wall temperature as Prw = 0.685 where this should be verified. Note that CC = m ˙ C c p,C = (12 kg/s)(1.0115 kJ/kg-K) = 12.1380 W/K and CH = m ˙ H c p,H = (12.93 kg/s)(1.0295 kJ/kg-K) = 13.3114 W/K which makes CC = Cmin and C∗ =

12.1380 W/K = 0.9118 13.3114 W/K

Moreover, the required heat exchanger effectiveness will be, as given by Equation 25.23 =

CC (t2 − t1 ) t2 − t1 204◦ C − 0◦ C = = = 0.6375 Cmin (T1 − t1 ) T1 − t1 320◦ C − 0◦ C

and the value of Ntu required to yield this effectiveness may be obtained from Equation 25.36 1 1 − C ∗ ln 1 − C∗ 1− 1 1 − (0.6375)(0.9118) = ln 1 − 0.9118 1 − 0.6375 0.4187 = 11.344 ln 0.3625 = 11.344 ln 1.1550 = (11.344)(0.1441) = 1.6349

Ntu =

The total number of tubes will be nt = (48)(14) = 672 tubes and Table A.21 reveals that the tubes have inner and outer diameters of di = 4.526 cm and

do = 5.080 cm

which makes the required UC SC product UC SC = 19.844 w/k. In the first trial, we will assume a total length of 6 m and for tubes spaced on 7.5 cm centers (both ST = SL = 7.5 cm), the air preheater width will be W = (48)(0.075 m) = 3.60 m

832

Introduction to Thermal and Fluid Engineering

The free-flow area on the tube (flue gas) side will be   AH = nt di2 = (672)(0.04526 m) 2 = 1.0812 m2 4 4 and with L p and L designated as the tube length per pass and the total tube length, respectively, we have the cold-side free duct area     L 6m AC = L p W = (3.60 m) = (3.60 m) = 4.32 m2 5 5 The two heat transfer coefficients may now be determined. For the hot (flue gas) side m ˙H 12.93 kg/s Vˆ H = = = 16.97 m/s  H AH (0.7048 kg/m3 )(1.0812 m2 ) and Red =

h Vˆ h di (0.7048 kg/m3 )(16.97 m/s)(0.04526 m) = = 20,270 h 2.671 × 10−5 kg/m-s

which shows that the flue gas on the inside of the tubes is in turbulent flow. We therefore employ Equation 22.18 with the exponent on Pr taken as 0.30 because Tw < Tb . 0.30 Nud = 0.023Re0.80 = 0.023(20, 270) 0.80 (0.680) 0.30 = 57.14 d Pr

so that hH

k = Nud = d



 0.0404 W/m-K (57.14) = 51.00 W/m2 -K 0.04526 m

For the cold (air) side, the free duct velocity is ˆ∞ = V

m ˙c 12.00 kg/s = = 2.954 m/s c Ac (0.9403 kg/m3 )(4.32 m2 )

and by Equation 23.18a     ST 7.5 cm ˆ∞ Vˆ max = V = (2.954 m/s) = 9.155 m/s ST − do 7.5 cm − 5.080 cm Then Red,max =

c Vˆ max do (0.9403 kg/m3 )(9.155 m/s)(0.0508 m) = = 20, 060 c 2.18 × 10−5 kg/m-s

and with C and m taken from Table 23.2 C = 0.27 and

m = 0.63

we have from Equation 23.17  Nud,max = 0.27(Red,max ) 0.63 (Pr) 0.36

Pr Prw 

= 0.27(20, 060) 0.63 (0.693) 0.36

1/4

0.693 0.685

1/4 = 121.83

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833

and hC =

k Numax = d



 0.0318 W/m-K (121.83) = 76.26 W/m2 -K 0.0508 m

With both h H and h C in hand, we may form the overall heat transfer coefficient. With Cc = Cmin , we use Equation 25.10 with SC do L do 5.080 cm = = 1.122 = = SH di L di 4.526 cm we have UC =

1 1 = SC /SH 1 1.122 1 + + 2 hH hC 51.00 W/m -K 76.26 W/m2 -K

=

1 1 = 0.0220 m2 -K/W + 0.0131 m2 -K/W 0.0351 m2 -K/W

=

1 = 28.48 W/m2 -K 0.0351 m2 -K/W

With SC = nt do L = (672)(0.0508 m)(6.00 m) = 643.47 m2 the UC SC product will be UC = (28.48 W/m2 -K)(643.48 m2 ) = 18.325 kW/K which falls short of the required value of 19.829 kW/K. The difference between the calculated and required values of the UC SC product requires that a trial and error be employed. Table 25.3 summarizes the procedure. It may be noted that the computations for L = 6 m are repeated in the table. TABLE 25.3

Summary of Trial-and-Error Procedure for Design Example 17 Assume L, m SC , m2 L p, m AC , m2 , ˆ ∞ , m/s V Vˆ , m/s max

Red,max Nud,max h C , W/m2 -K h H , W/m2 -K SC /SH h H , m2 -K/W 1/ h C , m2 -K/kW Sum UC = 1/sum, W/m2 -K UC SC , kW/K OK? Remedy

6.00

6.40

6.75

643.47 1.20 4.320 2.954

664.23 1.28 4.608 2.769

718.14 1.34 4.824 2.568

9.155 20,060 121.83 76.26 51.00 0.0220 0.0131 0.0351 28.47 18.325 No Increase L

8.583 18,800 116.95 73.21 51.00 0.0220 0.0137 0.0357 28.01 19.225 No Increase L

7.959 17,440 111.54 69.83 51.00 0.0220 0.0143 0.0363 27.53 19.770 Yes Accept this result

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Introduction to Thermal and Fluid Engineering

Although slightly lower than required, we have accepted the result of L = 6.70 m because this is within any accuracy limitations.

25.5

Finned Heat Exchangers

Attention is now focused on the finned heat exchanger in which the surface area on the hot and/or cold side of the exchanger is augmented with fins. Three such exchangers are shown in Figure 25.11 and the performance of these exchangers is governed by the enthalpy balance in Equation 25.1 ˙ = C H (T1 − T2 ) = CC (t2 − t1 ) Q

(25.1)

(a)

(b)

(c) FIGURE 25.11 Examples of finned heat exchangers. (a) The double pipe with the outside surface of the inner pipe containing longitudinal fins of rectangular profile, (b) a single tube in a tubular heat exchanger with radial or annular fins of a constant cross section, and (c) the compact heat exchanger with rectangular plate fins on both hot and cold sides.

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835

and the rate equation of Equation 25.2 ˙ = U Sm = UH SH m = UC SC m Q

(25.2)

25.5.1 The Surface Area and the Overall Surface Efficiency We first consider the outside surface of the inner tube or pipe in the double-pipe exchanger shown in Figure 25.11a. The heat exchange is between a fluid inside this inner pipe or tube and a fluid flowing in the annular region The outside surface of the inner tube or pipe contains n f longitudinal fins, usually fins of rectangular profile of height b and thickness . The entire exchanger has a length L and the diameter of the inner pipe is do . The base or prime surface area for the outside surface of the inner pipe will be Sb = (do − n f )L

(m)

(25.49a)

and because no heat flows from the tips of the fins, the finned surface is S f = 2n f b L

(m)

(25.49b)

The total surface area is So = Sb + S f = [do − n f  + 2n f b]L

(m)

(25.49c)

For the exchanger containing tubes with n f radial (also referred to as annular or circular) fins a similar procedure yields Sb = (L − n f )do

(m)

(25.50a)

where  is the fin thickness and do is the outside diameter of the tube. Because no heat flows from the tips of the fins, the finned surface is   S f = 2 da2 − do2 n f (m) (25.50b) 4 where da and do are, respectively, the outer and inner diameter of the fins. The total surface is   So = Sb + S f = (L − n f )do + 2 da2 − do2 n f (m) (25.50c) 4 There are several types of surfaces that can be used in the compact heat exchanger shown in Figure 25.11c. Typical of these are the plain plate fin, louvered fin, strip fin, and wavy fin surfaces as described by Kays and London (1984). Each surface possesses its own geometric properties and heat transfer and flow friction data. The total surface is determined from the volume of the exchanger and the finned surface to total surface ratio is given. We have discussed the efficiency of extended surface or fins in Chapter 20. There it was shown that the efficiency of the longitudinal fin of rectangular profile or a spine of constant cross section is given by =

tanh mb mb

(20.67)

The graph of the efficiency of the annular or radial fin of the rectangular profile given as Figure 20.24 can be replaced by an approximate expression due to McQuiston and Tree (1972) =

tanh m m

(25.51)

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Introduction to Thermal and Fluid Engineering

where, in both Equation 20.67 and Equation 25.51  m=

2h k

1/2

and in Equation 25.51   = rb

1− 

  1 1 + 0.35 ln 

where  is the radius ratio,  = ro /ra ≤ 1. We now develop an expression for the overall surface efficiency. With the total surface equal to the sum of the base or prime surface and the finned surface, we may consider the prime surface at an efficiency of unity and the finned surface at its own fin efficiency. Hence, in ov So = (1.00)Sb + f S f we note that, with Sb = So − S f , we will have ov So = (1.00)(So − S f ) + f S f   Sf Sf ov = 1 − + f So So or ov = 1 − (1 − f )

Sf So

(25.52)

25.5.2 The Overall Heat Transfer Coefficient Several relationships for the overall heat transfer coefficient have been listed in Section 25.2.3; all were for the unfinned exchanger. If the exchanger contains fins, then the expressions for the overall heat transfer coefficient must include the effect of the overall fin efficiency. In the equations for the overall heat transfer coefficient that follow, no provision is made for fouling on the side of the exchanger that contains the fins. Such a provision is somewhat beyond our scope and information on this topic can be found in Kraus et al. (2001). •

For the hot fluid flowing on the finned side of the exchanger with fouling on the unfinned cold side and no metal thermal resistance UH =

•

1 h H ov, H

1 1 SH 1 SH + + h d,C SC h C SC

(25.53)

For the cold fluid flowing on the finned side of the exchanger with fouling on the unfinned hot side and no metal thermal resistance UC =

1 1 SC 1 SC 1 + + h H SH h d, H SH h C ov,C

(25.54)

Heat Exchangers •

837

For the hot fluid flowing on the finned side of the exchanger with no fouling and no metal thermal resistance UH =

•

(25.55)

1 1 SC 1 + h H SH h C ov,C

(25.56)

For the case of both sides of the exchanger finned with no fouling and no metal thermal resistance, we have for the hot-side reference UH =

•

1 1 SH + h H ov, H h C SC

For the cold fluid flowing on the finned side of the exchanger with no fouling and no metal thermal resistance UC =

•

1

1 1 1 SH ++ h H ov, H h C ov,C SC

(25.57)

For the case of both sides of the exchanger finned with no fouling and no metal thermal resistance, we have for cold-side reference UC =

1 1 SC 1 + h H ov, H SH h C ov,C

(25.58)

We now turn to an example that will illustrate the concepts discussed in this section.

Example 25.6 A double-pipe heat exchanger with a finned annulus is to be used in a liquid-to-liquid application. The hot liquid with a capacity rate of 2.275 kW/K and an inlet temperature of T1 = 85◦ C flows in the annulus in counterflow with a heat transfer coefficient of 2500 W/m2 -K. The cold liquid with a capacity rate of 2.844 kW/K and an inlet temperature of t1 = 25◦ C flows in the inner pipe with a heat transfer coefficient of 2000 W/m2 -K. The 3-in × schedule-40 inner pipe resides within a 5-in-schedule 40 pipe, and both pipes are 1.2-m long. The inner pipe is equipped with 8 longitudinal fins of rectangular profile, each 1.905-cm high and 0.3175-cm thick. The thermal conductivity of the inner pipe material is 42 W/m-K and fouling may be considered negligible. Determine the outlet temperatures of both fluids.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) There is negligible heat loss to and from the surroundings. (3) Kinetic and potential energy changes are negligible. (4) The tube wall thermal resistance is negligible. (5) The Murray-Gardner assumptions pertaining to fins listed in Chapter 20 apply. (6) The fins possess adiabatic tips. (7) Fouling of the exchanger surfaces is specified as being negligible.

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Introduction to Thermal and Fluid Engineering

(8) Fluid specific heats and the overall heat transfer coefficient remain constant. (9) Axial conduction along the piping is negligible. For the 3-in × schedule-40 inner pipe, Table A.20 gives do = 8.890 cm

di = 7.793 cm

and

We then make the observation that C H = 2.275 kW/K and CC = 2.844 kW/K so that C H = Cmin = 2.275 KW/K. The hot fluid flows in the annulus and only the surface area of the inner pipe need be obtained. Thus, with So = SH . The base surface is obtained from Equation 25.49a Sb = (do − n f )L = [(8.890 cm) − 8(0.3175 cm)](120 cm) = (27.929 cm − 2.540 cm)(120 cm) = 3046.7 cm2

or

0.3047 m2

and for the finned surface, Equation 25.49b gives S f = 2n f b L = 2(8)(1.905 cm)(120 cm) = 3657.6 cm or

0.3658 cm2

Then So = SH = Sb + S f = 0.3047 cm2 + 0.3658 cm2 = 0.6705 cm2 and Sf 0.3658 cm2 = = 0.546 So 0.6705 cm2 The fin efficiency for the hot side is given by Equation 20.67. We first find m 

2h m= k



1/2 =

2(2500 W/m2 -K) (42 W/m-K)(0.003175 m)

1/2 = 193.64 m−1

Then mb = (193.64 m−1 )(0.01905 m) = 3.689 and Equation 20.67 gives f =

tanh mb tanh 3.689 0.9988 = = = 0.271 mb 3.689 3.689

Then we use Equation 25.52 to find the overall passage efficiency for the annulus: ov, H = 1 − (1 − f )

Sf SH

= 1 − (1 − 0.271)(0.546) = 1 − (0.729)(0.546) = 0.602

Heat Exchangers

839

The surface area of the inside of the inner pipe will be needed for the determination of the overall heat transfer coefficient. Because the inside of the inner pipe carries the cold fluid, this will be designated as Si = SC SC = di L = (7.793 cm)(120 cm) = 2937.8 cm2

(0.2938 m2 )

We then use Equation 25.52 to determine the overall heat transfer coefficient UH =

1 1 1 SH + h H ov, H h C SC 1

=

1 2

(2500 W/m -K))(0.602)

+

0.6705 m2 2000 W/m -K 0.2938 m2 1

2

=

1 6.645 × 10−4 m2 -K/W + 1.141 × 10−3 m2 -K/W

=

1 = 553.9 W/m2 -K 1.806 × 10−3 m2 -K/W

Then, with C H = Cmin = 2.275 kW/K Ntu =

UH SH (553.9 W/m2 K)(0.6705 m2 ) = = 0.1632 CH 2.275 kW/K

and C∗ =

Cmin 2.275 kW/K = = 0.800 Cmax 2.844 kW/K

Equation 25.35a for the case of counterflow then gives ∗

=

1 − e −Ntu (1−C ) 1 − e −(0.1632(1−0.800) ∗) = ∗ −N (1−C tu 1−C e 1 − (0.800)e −0.1632(1−0.800)

=

1 − e 0.0326 1 − 0.9679 = 1 − (0.800)e −0.0326 1 − (0.800)(0.9679)

=

0.0321 = 0.142 0.2257

Then from Equation 25.33 with C H = Cmin =

T1 − T2 T1 − t1

T1 − (T1 − T1 ) = 85◦ C − (0.142)(85◦ C − 25◦ C) = 85◦ C − (0.142)(60◦ C) = 85◦ C − 8.52◦ C = 76.5◦ C ⇐ and because by definition with C H = Cmin C∗ =

Cmin CH t2 − t1 = = Cmax CC T1 − T2

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Introduction to Thermal and Fluid Engineering

we find that t2 = t1 + C ∗ (T1 − T2 ) = 25◦ C + (0.800)(85◦ C − 76.5◦ C) = 25◦ C + (0.800)(8.5) = 31.8◦ C ⇐

25.6

Summary

The duty of a heat exchanger is given by the enthalpy balance ˙ = C H (T1 − T2 ) = Cc (t2 − t1 ) Q

(25.1)

and whether this duty can be accomplished by an exchanger of a given size depends on the rate equation ˙ = U Sm = UH SH m = UC SC m Q

(25.2)

In Equation 25.2, the inside and outside surfaces for an unfinned tubular heat exchanger are given by Si = di L

and

So = do L

The overall heat transfer coefficient involves as many as five resistances, some of which may not be present. A consideration of all of the resistances except the metal resistance leads to U=

1 S S S S + + + h H SH h d, H SH h d,C SC h C SC

(25.12)

with other more specific cases given by Equations 25.13 through 25.16. For true counterflow, true cocurrent flow, or constant temperature source and receiver fluids, the logarithmic mean temperature difference can be evaluated from m = LMTD =

T1 − T2 T2 − T1 = T1 T2 ln ln T2 T1

(25.23)

For arrangements other than true counterflow, true cocurrent flow, or constant temperature source and receiver fluids, the LMTD correction factor method or the Ntu method can be employed. The LMTD correction factor, F , can be obtained from charts that consider two parameters P=

t2 − t1 T1 − t1

and

R=

T1 − T2 CC = t2 − t1 CH

with F = f ( P, R) obtained from charts such as those shown in Figures 25.6 through 25.9. With the F factor so obtained, the rate equation becomes ˙ = U SF (LMTDc ) Q where LMTDc is the logarithmic mean temperature difference for true counterflow.

Heat Exchangers

841

The Ntu method is based on three parameters C∗ =

Cmin , Cmax

=

˙ Q , ˙ max Q

and

Ntu =

US Cmin

The heat transfer effectiveness is a function of Ntu and C ∗ . Values have been provided for eight arrangements in Equations 25.35 through 25.48. In dealing with finned surfaces, care must be used to consider the fin efficency. Most often, an overall passage efficiency is employed ov = 1 − (1 − f )

Sf So

(25.52)

where So is the total surface and S f is the finned surface. In the overall heat transfer coefficient, the individual heat transfer coefficients must be multiplied by their respective values of ov .

25.7

Problems

The Heat Balance 25.1: Ethylene glycol, flowing at a rate of 200 kg/s, enters a heat exchanger at 80◦ C and leaves at 40◦ C. The glycol is cooled by water, flowing at 160 kg/s, that enters at 15◦ C. Determine (a) the heat transferred in the exchanger and (b) the exit temperature of the water. 25.2: Rework Problem 25.1 with the glycol cooled by an airflow of 1425 kg/s of air at 20◦ C. All other conditions remain the same. 25.3: An air-cooled power plant condenser handles 3.2 kg/s of steam that has a quality of 98% when it enters and is saturated liquid when it leaves. The condenser pressure is 10 kPa and air enters the condenser at 20◦ C at a flow rate of 540 kg/s. Determine (a) the heat transferred in the condenser and (b) the exit temperature of the air. 25.4: The steam in Problem 25.3 is condensed by a more conventional flow of river water that enters at 17◦ C at a flow rate of 135 kg/s. Determine the exit temperature of the water. 25.5: A stream of lubricating oil (SAE 50) at 70◦ C is cooled by water. The oil flows at a rate of 24 kg/s; the water, flowing at a rate of 20 kg/s, enters the exchanger at 15◦ C and leaves at 23◦ C. Determine (a) the heat transferred in the exchanger and (b) the exit temperature of the oil. 25.6: Rework Problem 25.5 using air at a flow rate of 8 kg/s that enters at 15◦ C and leaves at 33◦ C. All other conditions remain the same. 25.7: In a gas-to-gas application, nitrogen is cooled from 80◦ C to 30◦ C by air, entering at 15◦ C. The flow rate of both the nitrogen and the air is 10 kg/s. Determine (a) the heat transferred in the exchanger and (b) the exit temperature of the air. 25.8: Chilled water entering a heat exchanger at 8◦ C is used to cool air from 40◦ C to 12◦ C. The flow rates of the water and the air are 8 and 15 kg/s, respectively. Determine (a) the heat transferred in the exchanger and (b) the exit temperature of the water.

842

Introduction to Thermal and Fluid Engineering

25.9: Rework Problem 25.1 using glycerin. All other conditions remain the same. 25.10: Rework Problem 25.8 using carbon dioxide. All other conditions remain the same. The Overall Heat Transfer Coefficient 25.11: A hot liquid flows in the interior of 2-in × 12-BWG tubing with a heat transfer coefficient of 1250 W/m2 -K. A cold fluid flows on the outside of the tubing with a heat transfer coefficient of 600 W/m2 -K. Fouling may be neglected. Determine the overall heat transfer coefficient based on (a) the hot side and (b) the cold side. 25.12: Rework Problem 25.11 with a fouling resistance on the hot side of 1/ h d, H = 0.0002 m2 K/W. 25.13: Rework Problem 25.11 with a fouling resistance on the cold side of 1/ h d,C = 0.0002 m2 K/W. 25.14: Rework Problem 25.11 with fouling resistances on both the hot and cold sides of 1/ h d, H = 1/ h d,C = 0.0002 m2 -K/W. 25.15: A hot liquid flows in the interior of 2-in × 12-BWG tubing with a heat transfer coefficient of 800 W/m2 -K. A cold gas flows on the outside of the tubing with a heat transfer coefficient of 100 W/m2 -K. Fouling may be neglected. Determine the overall heat transfer coefficient based on (a) the hot side and (b) the cold side. 25.16: Rework Problem 25.15 with a fouling resistance on the hot side of 1/ h d, H = 0.0004 m2 K/W. 25.17: Rework Problem 25.15 with a fouling resistance on the cold side of 1/ h d,C = 0.0004 m2 -K/W. 25.18: Rework Problem 25.15 with the same fouling resistances on both the hot and cold sides: 1/ h d, H = 1/ h d,C = 0.0004 m2 -K/W. 25.19: A tubular air heater is constructed of low carbon steel tubing with a thermal conductivity k of 42 W/m-K. The tubes are arranged in a bank that is twelve tubes wide by ten tubes deep. The tubes are 2-m long and have an outside diameter of 7.62 cm and an inside diameter of 7.198 cm. The hot (inside) and cold (outside) heat transfer coefficients are 150 W/m2 -K and 100 W/m2 -K, respectively. Fouling may be considered negligible. Determine the overall heat transfer coefficient based on (a) the hot side, (b) the cold side, and (c) the hot side when the tube metal resistance is included. 25.20: Rework Problem 25.19 with a fouling resistance on the hot side of 0.0025 m2 -K/W. 25.21: Rework Problem 25.19 with a fouling resistance on the cold side of 1/ h dC = 0.0004 m2 K/W. 25.22: Rework Problem 25.19 with a fouling resistances on both the hot and cold sides of 1/ h d H = 1/ h dC = 0.0002 m2 -K/W. 25.23: The tubing in Problem 25.19 is used as an air-cooled steam condenser. The steam is condensed on the inside of the tubing with a heat transfer coefficient of 8000 W/m2 -K. Air with a heat transfer coefficient of 100 W/m2 -K is used on the outside of the tubes. Fouling may be neglected. Determine the overall heat transfer coefficient based on the hot side. 25.24: Rework Problem 25.23 to determine the overall heat transfer coefficient based on the cold side. 25.25: Rework Problem 25.23 for fouling resistances on the hot and cold sides of h d, H = 1/0.0020 m2 -K/W and 1/ h d,C = 0.0002 m2 -K/W, respectively.

Heat Exchangers

843

The Logarithmic Mean Temperature Difference 25.26: In a single-pass heat exchanger, the hot fluid enters at 300◦ C and leaves at 120◦ C. The cold fluid enters at 80◦ C and leaves at 140◦ C. Assume true counterflow and determine the LMTD. 25.27: In a single-pass counterflow heat exchanger, the hot fluid enters at 300◦ C and leaves at 140◦ C, and the cold fluid enters at 80◦ C and leaves at 120◦ C. Determine the LMTD. 25.28: In a single-pass cocurrent flow heat exchanger, the hot fluid enters at 300◦ C and leaves at 140◦ C, and the cold fluid enters at 80◦ C and leaves at 120◦ C. Determine the LMTD. 25.29: Steam at 5 kPa is condensed in a steam power plant in a condenser that takes cooling water from an adjacent river. The cooling water enters at 20◦ C and leaves at 24◦ C. Determine the LMTD. 25.30: Air enters the evaporator of an air-conditioning unit at 32◦ C and leaves at 22◦ C. The system employs refrigerant-134a at 2.8 bar in the evaporator coil. Determine the LMTD. 25.31: The surface of a metal plate used as a heat exchanger is held at a constant temperature. Air enters at 25◦ C and leaves at 60◦ C; the LMTD is 78◦ C. Determine the surface temperature. 25.32: In a counterflow heat exchanger, the hot stream is cooled from 340◦ C to 140◦ C by a fluid that is heated from 60◦ C to an unknown temperature, t2 . Determine t2 for an LMTD of 135◦ C. 25.33: In a cocurrent flow heat exchanger, the cold fluid is heated from 40◦ C to 70◦ C. The hot fluid enters at 260◦ C and leaves at an unknown temperature, T2 . Determine T2 for an LMTD of 86◦ C. The LMTD Correction Factor Method 25.34: A shell-and-tube heat exchanger is used for heating 12 kg/s of oil (c C = 2.2 kJ/kg-K) from 20◦ C to 38◦ C. The heat exchanger has one shell pass and two tube passes. Hot water (c H = 4.18 kJ/kg-K) enters the shell at 75◦ C and leaves the shell at 55◦ C. The overall heat transfer coefficient based on the outside surface of the tubes is estimated to be 950 W/m2 -K. Determine (a) the corrected LMTD and (b) the required surface area in the exchanger. 25.35: Consider the heat exchanger in Problem 25.34. After 4 years of operation, the outlet of the oil reaches only 30◦ C instead of 38◦ C with everything else being the same. Determine the fouling resistance on the oil side of the exchanger. 25.36: A shell-and-tube heat exchanger is used to cool oil (c H = 2.2 kJ/kg-K) from 110◦ C to 65◦ C. The heat exchanger has two shell passes and four tube passes. The coolant (c C = 4.20 kJ/kg-K) enters the shell at 20◦ C and leaves the shell at 42◦ C. For an overall hot-side heat transfer coefficient of 1200 W/m2 -K and an oil flow of 11 kg/s, determine (a) the coolant mass flow rate and (b) the required surface area in the exchanger. 25.37: In an engine oil cooler, oil (c H = 2.08 kJ/kg-K) enters at 58◦ C with a mass flow rate of 2 kg/s. The coolant (c C = 4.18 kJ/kg-K) enters the cooler at 12◦ C and leaves at 24◦ C with a mass flow rate of 1.66 kg/s. Assuming an overall heat transfer coefficient of 2800 W/m2 -K, determine the outlet oil temperature and the heat transfer surface area for (a) a parallel flow and (b) a cross-flow arrangement with both fluids unmixed.

844

Introduction to Thermal and Fluid Engineering

25.38: In a shell-and-tube heat exchanger, the hot fluid (c H = 3.798 kJ/kg-K) flows at the rate of 7 kg/s through eighty 2-cm OD tubes. The hot fluid experiences a temperature drop of 20◦ C. The cold fluid (c C = 4.18 kJ/kg-K) enters the shell at 12◦ C with a mass flow rate of 6.4 kg/s. The overall heat transfer coefficient based on the tube outside surface area is 680 W/m2 -K. The exchanger has one shell pass and two tube passes. Determine the length of each tube. 25.39: A finned tube cross-flow heat exchanger with both fluids unmixed is used to heat water (c C = 4.2 kJ/kg-K) from 20◦ C to 75◦ C. The mass flow rate of the water is 2.5 kg/s. The hot stream (c H = 1.2 kJ/kg-K) enters the heat exchanger at 280◦ C and leaves at 120◦ C. The overall heat transfer coefficient is 130 W/m2 -K. Determine (a) the mass flow rate of the hot stream and (b) the exchanger surface area. 25.40: A cross-flow heat exchanger is used to preheat combustion air to a boiler from 20◦ C to 275◦ C by using hot exhaust gases at 780◦ C. The overall heat transfer coefficient is 150 W/m2 -K. The air flowing on the unmixed side of the exchanger has a mass flow rate of 17.31 kg/s with a specific heat of c p,C = 1.0167 kJ/kg-K. The exhaust gases flowing on the mixed side of the exchanger has a mass flow rate of 16.60 kg/s and a specific heat of c p, H = 1.1020 kJ/kg-K. Determine the required heat transfer area. 25.41: A steam condenser consists of a bundle of brass tubes housed in a shell. Each tube has an inside diameter of 2 cm. The cooling water (c C = 4.18 kJ/kg-K), which flows through the tubes enters the condenser at 20◦ C and leaves at 29◦ C with a mass flow rate of 1200 kg/s. Steam enters the condenser as saturated vapor at 15 kPa, condenses on the outside of the tube bundle and emerges from the condenser as saturated liquid at 15 kPa. The overall heat transfer coefficient based upon the outside tube surface area is 4800 W/m2 -K. Determine (a) the condensation rate of the steam in kg/s and (b) the number of tubes per pass for a two-tube pass arrangement with each tube 2.6-m long. 25.42: A steam condenser has brass tubes with a 1.9-cm inside diameter, a wall thickness of 4 mm, and a length of 3 m. The condenser is of a one shell pass-two tube pass design with 120 tubes per pass. Cooling water (c C = 4.18 kJ/kg-K) enters the tubes at 20◦ C with a velocity of 1.5 m/s. The heat transfer coefficients on the inside and outside of the tubes are 6500 W/m2 -K and 10,000 W/m2 -K, respectively. The pressure of the condensing steam is 10 kPa. Neglecting any conduction or fouling resistances, determine (a) the temperature of the water leaving the condenser and (b) the rate of steam condensation. 25.43: In a water-to-water shell and tube heat exchanger containing one shell pass and two tube passes, hot water flows on the shell side at 3.2 kg/s with an inlet temperature of 110◦ C. Cold water flows inside the tubes at a rate of 4.5 kg/s with an inlet temperature of 25◦ C and an outlet temperature of 60◦ C. The overall heat transfer coefficient based on the outside surface area of the tubes is 1300 W/m2 -K. Determine (a) the hot water exit temperature and (b) the surface in the exchanger. 25.44: In a shell and tube heat exchanger, glycerin flows through the 2-in × 10-BWG tubes and is heated from 22◦ C to 36◦ C. Hot water enters the shell side at 85◦ C and exits at 50◦ C providing a heat transfer coefficient of 350 W/m2 -K. There are a total of 80 tubes, each carrying glycerin at 0.6 kg/s. The inside and outside tube diameters are 5.00 cm and 5.08 cm, respectively. The heat transfer coefficient on the inside of the tubes may be taken as 400 W/m2 -K and the specific heats of water and glycerin are

Heat Exchangers

845

4.185 kJ/kg-K and 2.39 kJ/kg-K, respectively. Determine (a) the mass flow rate of the water, (b) the corrected LMTD, and (c) the tube length if each tube makes four passes. 25.45: Determine the heat transfer duty of the following heat exchanger arrangements if T1 = 150◦ C, T2 = 90◦ C, t1 = 25◦ C, t2 = 75◦ C, and U S = 130 W/◦ C. (a) Parallel flow double pipe. (b) Counterflow double pipe. (c) One shell pass-two tube passes. (d) Two shell passes-four tube passes. (e) Cross flow—both fluids unmixed. (f) Cross flow—one fluid mixed, the other fluid unmixed. 25.46: In a cross-flow exchanger with both fluids unmixed, the temperatures of the hot and cold fluids are T1 = 135◦ C, T2 = 50◦ C, t1 = 30◦ C, t2 = 68◦ C. The total hot-side heat transfer surface area provided by the heat exchanger is 10.5 m2 . The heat exchanger is to meet a transfer duty of 210 kW. Determine (a) the overall heat transfer coefficient referred to the hot side and (b) the ratio, C H /CC . 25.47: In a one shell pass-two tube pass heat exchanger, cooling water (c C = 4.18 kJ/kg-K) enters the shell side at 20◦ C and a flow rate of 7000 kg/h. Hot oil (c H = 2.095 kJ/kg-K) flows through the tubes, entering at 110◦ C and flowing at a rate of 14,000 kg/h. The total heat transfer area is 15.4 m2 and the overall heat transfer coefficient is 450 W/m2 -K with both referred to the hot side of the exchanger. Determine (a) the outlet temperatures of both the water and the oil and (b) the heat transfer duty of the exchanger. 25.48: Determine the hot-side heat transfer surface area for each of the following heat ˙ = exchanger arrangements if T1 = 150◦ C, T2 = 85◦ C, t1 = 30◦ C, t2 = 80◦ C, Q 2 180 kW, and UH = 350 W/m -K : (a) Parallel flow double pipe. (b) Counterflow double pipe. (c) One shell pass-two tube passes. (d) Two shell passes-four tube passes. (e) Cross flow—both fluids unmixed. (f) Cross flow—one fluid mixed, the other fluid unmixed. 25.49: Hot fluid flows through the tubes in a cross-flow heat exchanger in which the hotside heat transfer surface is 16 m2 . The inlet and outlet temperatures of the hot fluid are 120◦ C and 65◦ C, respectively. The cold fluid, assumed to be mixed, enters the heat exchanger at 25◦ C and leaves at 65◦ C. The heat transfer duty is 1000 W/m2 . Determine the overall heat transfer coefficient. 25.50: In a cross-flow heat exchanger, hot exhaust gases (c p, H = 1.045 kJ/kg-K), assumed to be unmixed, flow over a bank of tubes at the rate of 6 kg/s. The inlet and outlet temperatures of the exhaust gases are 380◦ C and 180◦ C, respectively. Water enters through the tubes as saturated liquid at 0.60 MPa and emerges as saturated vapor at the same pressure. The overall hot-side heat transfer coefficient is 210 W/m2 -K. Determine (a) the surface area in the exchanger and (b) the steam generation rate.

846

Introduction to Thermal and Fluid Engineering

25.51: Show that the logarithmic mean temperature difference correction factor, F , for a cocurrent (parallel) flow exchanger may be expressed as 1− P ln R+1 1 − PR F = 1 R−1 ln 1 − P( R − 1) were P and R are as defined in the text. 25.52: In a steam condenser, saturated steam enters the shell at 10 kPa and condenses into saturated liquid at the same pressure. Water enters the tubes at 25◦ C and flows at the rate of 200 kg/s. The mass of steam condensed is 6 kg/s. Determine (a) the outlet temperature of the water, (b) the hot-side heat transfer area if the condenser is in a one shell-two tube pass arrangement with an overall hot-side heat transfer coefficient of UH = 2500 W/m2 -K. 25.53: Consider the steam condenser in Problem 25.52. With the mass flow rates and the inlet and outlet temperatures fixed, determine the mass of steam condensed if the pressure of the steam is 15 kPa. The -Ntu Method 25.54: Show that for a cocurrent (parallel flow) exchanger, the -Ntu relationship is =

1 − e −Ntu (1+C 1 + C∗

∗

)

where C ∗ = Cmin /Cmax . 25.55: Considering the expression derived in Problem 25.54, show that for an infinitely long, cocurrent (parallel) flow heat exchanger, the expression reduces to =

1 1 + C∗

25.56: Using the equations ˙ = C H (T1 − T2 ) = CC (t2 − t1 ) Q and ˙ = U S (T1 − t2 ) − (T2 − t1 ) Q T1 − t2 ln T2 − t1 derive the -Ntu relationship for the counterflow heat exchanger ∗

=

1 − e −Ntu (1−C ) 1 − C ∗ e −Ntu (1−C ∗ )

where C ∗ = Cmin /Cmax . 25.57: Show that, in the limit as C ∗ −→ 1, the -Ntu expression derived in Problem 25.56 reduces to =

Ntu 1 + Ntu

Heat Exchangers

847

25.58: The temperature of the hot fluid in a heat exchanger drops from 75◦ C to 55◦ C while the cold fluid increases in temperature from 25◦ C to 35◦ C. Determine (a) the capacity rate ratio, C ∗ and (b) the number of transfer units, Ntu , if the exchanger is in counterflow, and (c) the number of transfer units, Ntu , if the exchanger is in cocurrent (parallel) flow. 25.59: The temperature of the hot fluid in a heat exchanger decreases from 68◦ C to 48◦ C while the cold fluid increases in temperature from 24◦ C to 36◦ C. Determine (a) the capacity rate ratio, C ∗ , (b) the number of transfer units, Ntu , if the exchanger is in cross flow with both fluids mixed, and (c) the number of transfer units, Ntu , if the exchanger is in cross flow with both fluids unmixed. 25.60: A double-pipe exchanger is used to cool oil (c H = 3.20 kJ/kg-K) from 110◦ C to 65◦ C. Water (c C = 4.20 kJ/kg-K) is used as the coolant entering at 20◦ C and leaving at 42◦ C. For an overall heat transfer coefficient of UH = 1000 W/m2 -K and an oil flow of 1.125 kg/s, determine (a) the coolant flow rate, (b) the required surface for a counterflow arrangement, and (c) the required surface for a cocurrent (parallel) flow arrangement. 25.61: Rework Problem 25.37 using the -Ntu method of analysis to determine the heat transfer area. 25.62: Rework Problem 25.39 using the -Ntu method of analysis to determine the heat transfer area. The heat transfer coefficient is referred to the hot side. 25.63: Rework Problem 25.40 using the -Ntu method of analysis to determine the heat transfer area. 25.64: A double-pipe exchanger, operating in counterflow, has the same operating conditions as in Problem 25.34. The hot water is in the outer pipe and the cold water is in the inner pipe. The overall heat transfer coefficient is based on the hot side. Use the -Ntu analysis method to find the heat transfer area. 25.65: A double-pipe exchanger, operating in counterflow is used as a clean water chiller. Clean water enters the annulus at 80◦ C and leaves at 40◦ C at a flow rate of 8 kg/s. Cold water enters the inner pipe at 20◦ C at a flow rate of 10 kg/s. The overall heat transfer coefficient is UH = 3450 W/m2 -K. Determine (a) the duty of the exchanger and (b) the surface required. 25.66: The water chiller in Problem 25.65 has the conditions of Problem 25.65 but the configuration contains 360 tubes that are 2 in × 10 BWG and are 80-cm long. The hot water flows on the outside of the tubes with a heat transfer coefficient of 4200 W/m2 -K and the cold water flows on the inside of the tubes with a heat transfer coefficient of 4800 W/m2 -K. It is typical of this cross-flow arrangement that both fluids are unmixed. Determine (a) the exchanger effectiveness and (b) the outlet temperature of the cold water.

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26 Radiation Heat Transfer

Chapter Objectives •

To consider electromagnetic waves and the electromagnetic spectrum and to show that thermal radiation is a form of electromagnetic radiation.

To consider the emission of radiant energy and the Stefan-Boltzmann and Wien displacement laws. • To define the terminology peculiar to radiation heat transfer such as emissivity, absorptivity, reflectivity, and transmissivity. • To describe what is meant by a blackbody and a gray body. •

•

To examine the directional characteristics of surface radiation.

To provide a relationship for radiant heat interchange with perfect absorbers and emitters and for surfaces not in full view of each other. • To modify the relationship for radiant heat interchange with nonperfect absorbers and emitters and for surfaces that are not in full view of each other. • To present an electrothermal analog method for handling radiation inside enclosures. •

26.1

The Electromagnetic Spectrum

Radiation of thermal energy is believed to be a specific form of radiation within the general phenomenon of electromagnetic radiation. As but one of numerous electromagnetic phenomena, thermal radiant energy travels at the speed of light: 2.9979 × 108 m/s. The existence of radiation as a mode of heat transfer can be observed from everyday experience. Consider, for example, a warm body enclosed without physical contact inside a cooler enclosure under complete vacuum. The warm body will eventually attain the temperature of the surrounding enclosure without the aid of conduction or convection. This statement may appear intuitive, and we can easily imagine, as well, the approximation of the warm body suspended by nonconducting cords in the evacuated cooler enclosure. All bodies continuously emit radiation. Figure 26.1 displays the electromagnetic spectrum showing a range of electromagnetic waves from long radio waves to the shorter wavelengths. The unit of wavelength used in this figure is the micrometer ( m) or 10−6 m,

849

850

Introduction to Thermal and Fluid Engineering Thermal Radiation 1μ

3

2

log λ, m 1 0

–1

–2

–3

–4

–5

–6

1A

–7

–8

–9

–10

–11

–12

X-rays Radio Waves

Infrared

Ultraviolet

γ-rays

Visible

FIGURE 26.1 The electromagnetic spectrum.

and the relationship between the frequency of this radiation, , and the wavelength, , is =

c 

(26.1)

where c is the speed of light. In Figure 26.1 we may note the very narrow band of wavelengths designated as the visible region. This band from 0.38 to 0.76 m is sensed by the optic nerve and is what we refer to as light. The radiation between 0.1 and 100 m is termed thermal radiation and we note that thermal radiation includes portions of the ultraviolet and infrared portions of the electromagnetic spectrum.

26.2

Monochromatic Emissive Power

Monochromatic emission refers to electromagnetic radiation having wavelengths confined to an extremely narrow wavelength range. In the present context, monochromatic emission may also be defined as hemispherical or spectral emission. 26.2.1 The Black Surface or Blackbody An ideal surface that absorbs all (and therefore reflects none) of the radiation falling upon it, regardless of the wavelength or the angle of incidence is defined as a black surface or blackbody. Because we “see” reflected light (in reality, electromagnetic radiation), the socalled blackbody, from which no light is reflected, will appear black. This is the idea behind the color black and we are quick to point out that surface color is not a requisite to the terminology. Because of the lacquer that it contains, black automobile paint does not yield a black surface. Moreover, freshly fallen snow, because of the diffuse nature of its reflection, may come very close to providing a black surface. A black surface emits a spectrum of radiant energy and this is considered in Section 26.2.2.

Radiation Heat Transfer

851

26.2.2 Planck’s Law Max Planck introduced the quantum concept in 1900 and, with it, the idea that radiation is emitted, not in a continuous state, but in discrete amounts or quanta. Planck’s law for the monochromatic emissive power of a surface is given as 2C1 c 2

E b (, T) =

5 (e cC1 /C2 T − 1)

(W/m2 )

(26.2)

where, in addition to c, which is the speed of light, the accepted values of the two constants are C1 = 6.624 × 10−27 erg which is Planck’s constant and C2 = 1.380 × 10−16 erg/K which is Boltzmann’s constant. We may note that while Planck’s law does indeed show that the monochromatic emissive power is a function of temperature and wavelength, it does not tell us how much radiation will occur at a given temperature. This is obtained from its integration over all wavelengths. 26.2.3 Wien’s Displacement Law Figure 26.2 indicates that there will be some value of wavelength for a particular absolute temperature that will yield a maximum value of E b (, T). A relationship between the maximum value of the product T can be obtained at the point where d E b (, T)/d vanishes. First, however, we make the transformation   A C1 c x= A=  C2 T so that =

A x

and

dx A =− 2 d 

Then    d E b (, T) d E b (, T) d x d (2C1 c 2 /A5 )x 5 A = = − =0 d dx d dx ex − 1 2 and the differentiation and algebraic expansion of terms can proceed to give 5(e x − 1) = xe x   C1 c C1 c/C2 T 5 e C1 c/C2 T − 1 = e C2 T which will be satisfied when T = 2897.8 -K

(26.3)

852

Introduction to Thermal and Fluid Engineering 108

107

106

105

104

103

102

Hemispherical Spectral Emissive Power Eλb (λ, T), Btu/(h . ft2 . μm)

Hemispherical Spectral Emissive Power Eλb(λ, T), W/(m2 . μm)

108 107

106 Black-body temperature T, ºR (K) 10,000 (5555) 5,000 (2778)

105

104

3,000 (1667)

103

1,500 (833) 102

1,000 (555) Locus of maximum values, Eλb (λmax, T)

101

101 100

0 Violet

2

4 6 8 Wavelength λ, μm Visible region (0.4 – 0.76 μm)

10

12

Red

FIGURE 26.2 Monochromatic emissive power as a function of wavelength and temperature.

Equation 26.3 is Wien’s1 displacement law which provides the wavelength for the maximum monochromatic emissive power for the temperature specified. 26.2.4 The Stefan-Boltzmann Law The total quantity of blackbody radiation emitted at a particular temperature can be obtained from an integration of Planck’s law over all wavelengths  Eb =

∞

 E b (, T)d =

0

∞

2C1 c 2

0

−5 d −1

e C1 c/C2 T

This time we make the transformation z=

1 W.

A C1 c/C2 =  T

Wien (1864–1928) was a German physicist who formulated the two Wien laws pertaining to radiation from blackbodies.

Radiation Heat Transfer

853

so that =

A z

and

d = −

A dz z2

With the substitution of the transformed variable we obtain  ∞ z3 dz E b = −2C1 c 2 A4 (e z − 1) 0 However, 1/(e z − 1) can be written as an infinite series so that a term-by-term integration can be performed to yield, after a substitution of the limits    4 4 C2 T 2C1 c 2 3! 3! 3! 3! 2 Eb = + 4 + 4 + 4 + · · · = 2c (6.45) A4 14 2 3 4 C12 c 4 or Eb =

  2(6.45)C24 2C1 c 2 (6.45) = T4 A4 C13 c 2

where the bracketed term, designated by , is a constant. Hence, E b = T 4

(26.4)

which is the Stefan-Boltzmann law. In Equation 26.4, the constant, , is universally designated as the Stefan-Boltzmann constant and has the value  = 5.67 × 10−8 W/m2 -K4 and we also observe that the Stefan-Boltzmann law can be written as E b = 5.67 × 10−8 T 4

(26.5)

We are frequently interested in the quantity of emission in a specific portion of the wavelength spectrum. This may be conveniently expressed by a fraction of the total emissive power and is designated by F1 -2 2 2 1 E b (, T)d  E b (, T)d = 1 F1 -2 = ∞ T 4 0 E b (, T)d or with both sides divided by T 5 F1 -2 =

 2 1 E b (T)d(, T) T 5 0   1 − E b (, T)d(T) = F0-2 − F0-1 0

We see that at a given temperature, the fraction of the emission between the two wavelengths of interest may be obtained from a subtraction of values. This subtraction process may be simplified if the variable of integration is changed to the wavelength-temperature product, T  2 T  1 T E b (d, T) E b (T) F1 -2 = d(T) − d(T) 5 T T 5 0 0

854

Introduction to Thermal and Fluid Engineering TABLE 26.1

Radiation Functions T m-K

F0-λT

Ebλ (λT) (cm-K)−1 T 5

λT m

F0-λT

Ebλ (, T) (cm-K)−1 σT 5

1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 2897.8 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000 5200 5400 5600 5800 6000

0.0003 0.0021 0.0078 0.0197 0.0396 0.0667 0.1009 0.1402 0.1831 0.2279 0.2501 0.2732 0.3181 0.3617 0.4036 0.4434 0.4809 0.5160 0.5488 0.5793 0.6075 0.6337 0.6579 0.6803 0.7010 0.7201 0.7378

0.0732 0.0855 0.1646 0.7825 1.1797 1.5499 1.8521 2.0695 2.2028 2.2623 2.2688 2.2624 2.2175 2.1408 2.0429 1.9324 1.8157 1.6974 1.5807 1.4679 1.3601 1.2590 1.1640 1.0756 0.9938 0.9181 0.8485

6200 6400 6600 6800 7000 7500 8000 8500 9000 9500 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 25,000 30,000 35,000 40,000 45,000 50,000

0.7541 0.7692 0.7832 0.7961 0.8081 0.8344 0.8562 0.8736 0.8900 0.9030 0.9142 0.9318 0.9451 0.9551 0.9628 0.9689 0.9738 0.9777 0.9808 0.9834 0.9856 0.9922 0.9953 0.9970 0.9979 0.9985 0.9989

0.7844 0.7255 0.6715 0.6220 0.5765 0.4786 0.3995 0.3354 0.2832 0.2404 0.2052 0.1518 0.1145 0.0878 0.0684 0.0540 0.0432 0.0389 0.0285 0.0235 0.0196 0.0087 0.0044 0.0025 0.0015 0.0009 0.0004

or F1 -2 = F0-2 T − F0-1 T

(26.6)

Values of F0-T are given in Table 26.1.

Example 26.1 The surface of a blackbody is held at 800 K. Determine (a) the total emissive power, (b) the wavelength at which the maximum monochromatic emissive power occurs, (c) the magnitude of the monochromatic emissive power at this wavelength, and (d) the fraction of the total emission between 3 m and 4 m.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The surface emits as a blackbody. (3) Radiation is the only mode of heat transfer. (a) At T = 800 K, Equation 26.4, which is the Stefan-Boltzmann law, gives E b = T 4 = (5.67 × 10−8 W/m2 -K4 )(800 K) 4 = 23.22 kW/m2 ⇐

Radiation Heat Transfer

855

(b) Equation 26.3, which is the Wien displacement law, yields =

2897.8 m-K = 3.622 m ⇐ 800 K

(c) At 3.622 m and 800 K where T = 2897.8 m-K, Table 26.1 gives E b (, T)/T 5 = 2.2688, (cm-K) −1 Hence, E b (, T) = 2.2688 × 10−4 (cm-K−1 )T 5 E b (3.622 m, 800 K) = [2.2688 × 10−4 (cm-K−1 )](5.670 × 10−8 W/m2 -K4 )(800 K) 5 = 4215.3 W/m2 -m ⇐ (d) The fraction of the total emission that occurs between 1 = 3 m and 2 = 6 m is obtained from Table 26.1. At 1 T = (3 m)(800 K) = 2400 m-K, read F0-2400 = 0.1402 and at 2 T = (6 m)(800 K) = 4800 m-K, read F0-4800 = 0.6075 Thus, F6.0-3.0 = F0-4800 − F0-2400 = 0.6075 − 0.1402 = 0.4673

26.3

(46.73%) ⇐

Radiation Properties and Kirchhoff’s Law

26.3.1 Absorptivity, Reflectivity, and Transmissivity Consider a system (or body) that possesses an element of surface area, dS, as shown in Figure 26.3a. We define the irradiation, G, as the total radiation heat flux that is intercepted by dS from all other surfaces that can “see” dS and we note that, in general, G may possess: •

An absorbed portion, G  = G, which takes place within the molecular layer immediately below the element of surface, dS.

•

A reflected portion, G  = G, which is reflected back into the entire space that can be “seen” by dS. Here we note that the reflected radiation may leave dS in all directions (Figure 28.3b) with no direction preferred. This type of radiation is referred to as diffuse and it is the type of radiation most likely to occur with rough surfaces. On the other hand, if the radiation reflected from the surface is directional such that the angle of incidence is equal to the angle of reflection (Figure 26.3c), the radiation is termed specular and is most likely to occur on smooth polished surfaces. The development here considers only diffuse radiation.

•

A transmitted portion, G = G, which can pass through the body and exit at some surface remote from dS.

856

Introduction to Thermal and Fluid Engineering G All Directions

Reflected Rays All Directions

dS

G

θ1

θ2

ρG

G αG

θ1 = θ2

τG (a)

(b)

(c)

FIGURE 26.3 (a) A small areas dS subjected to a total irradiation G, (b) the reflection pattern for diffuse radiation, and (c) the reflection pattern for specular radiation.

Conservation of energy may be applied to the very thin disk-shaped control volume that encloses dS in Figure 26.3a to show that with G  + G  + G = G G + G + G = G or ++ =1

(26.7)

where , , and are nonnegative dimensionless numbers that are system properties and we observe that none of them can exceed unity. Here  is the total absorptivity,  is the total reflectivity, and is the total transmissivity of the material. Because G is the irradiation from all directions over a hemispherical space above dS over all wavelengths, these properties are referred to as total hemispherical properties. If the surface is opaque, = 0 and Equation 26.7 reduces to +=1

( = 0)

(26.8)

If, in addition, the reflectivity is zero, the surface is black and the solid is said to be a black body. In this case, =1

( = 0,  = 0)

(26.9)

26.3.2 Kirchhoff’s Radiation Law Consider Figure 26.4 and let the temperatures of the black enclosure and the small nonblack test body be, respectively, designated as Tb and T1 . Further, let blackbody radiation impinge on the smaller body so that the amount of heat absorbed by the smaller body is equal to 1 E b . However, the smaller body will also emit radiation, say E 1 , and we note that the net rate of radiant heat interchange between the enclosure and the small body will be q˙ = 1 E b − E 1

Radiation Heat Transfer

857 Small test body (nonblack)

Blackbody enclosure at temperature Tb FIGURE 26.4 A small nonblack test body within a large black enclosure.

At thermal equilibrium when T1 = Tb , there can be no radiant heat interchange and q˙ = 0. Hence, 1 E b − E 1 = 0 or

Eb =

E1 1

and we can derive a similar relationship for n small nonblack bodies in thermal equilibrium Eb =

E1 E2 En = = ··· = 1 2 n

(26.10)

The general relationship of Equation 26.10 is known as Kirchhoff’s law.2 26.3.3 Emissivity All blackbodies at the same temperature will emit radiation at the same rate. If the smaller body in Figure 26.4 is a blackbody, then 1 = 1 and E 1 = E b . If, for the sake of argument, we propose that E 1 > E b , the smaller body must be cooled by a transfer of heat from the lower-temperature smaller body to the higher-temperature blackbody. Clearly, this would be in violation of the Second Law of Thermodynamics, which prohibits heat from flowing of its own accord from a region at low temperature to one at high temperature. It is also obvious from the foregoing that, if the smaller body is a blackbody, at thermal equilibrium it will emit exactly the same amount of radiation as the blackbody in which it is enclosed. The foregoing reasoning leads to three conclusions: 1. For thermal equilibrium, Kirchhoff’s law shows that the ratio of the emissive power of the surface to its absorptivity is the same for all bodies. 2. Because the absorptivity can never exceed unity at a particular temperature, a blackbody has the maximum emissive power of any surface. 3. The blackbody may be considered as a perfect emitter as well as a perfect absorber of radiant energy. In addition, because a perfect emitter or absorber does not exist, the concept of a blackbody is an idealized one. Indeed, the emissive characteristics of the so-called blackbody may be quite different from its absorptive characteristics. The emissive characteristics at a particular temperature are represented by the property known as the emissivity, , defined as the ratio of the actual rate of energy emission to the rate of energy emission of a blackbody 2 G.

R. Kirchhoff (1824–1887) was a German physicist who formulated the laws of electric currents and the electromotive forces in an electrical network and with R. W. Bunsen who developed the method of spectrum analysis.

858

Introduction to Thermal and Fluid Engineering

at the same temperature. Thus, E = E b Consider Figure 26.4 once again and assume that the smaller body is not a blackbody. We see that for a black enclosure with energy emitted, E b , the energy absorbed by the smaller body is E 1 = 1 E b and that the energy emitted by the smaller body is E 1 = 1 E b as required by the definition of emissivity. The net rate of heat interchange between the enclosure and small body is q˙ = 1 E b − 1 E b and when the small body and enclosure are at the same temperature (q˙ = 0), 1 E b = 1 E b from which we observe that the absorptivity equals the emissivity 1 = 1 26.3.4 An Approximation to a Blackbody Although no body or surface is a perfect emitter or blackbody, such a body may be approximated by a small hole cut into a hollow cavity. Suppose that the cavity in Figure 26.5 has its interior wall at a constant temperature, that is, let the interior wall of the cavity be an isothermal surface. The wall, of course, is nonblack and has an emissivity, . Now imagine that a small portion of the surface emits energy, E b , and that this energy escapes through the hole. Next, consider another portion of the surface that emits energy,

E b , and let this energy be reflected once before it escapes through the hole. This oncereflected energy will be equal in magnitude to  E b . For another portion of the surface that emits energy that is reflected twice, the magnitude of the escaping energy will be 2 E b . We see that the total energy emitted from the hole will contain energy that has been directly ρεEb

εEb

ρ2εEb FIGURE 26.5 Cavity used to produce blackbody radiation.

Radiation Heat Transfer

859

radiated, reradiated once, twice, and so on. The total escaping energy will be equal to E T = E b +  E b + 2 E b + 3 E b + · · · or E T = (1 +  + 2 + 3 + · · · ) E b However, we know that at any temperature,

==1− and we may take note of the infinite series 1 = 1 +  + 2 + 3 + · · · 1− Therefore, we see that the total energy can be represented by ET =

E b 1−  = Eb 1− 1−

or E T = Eb This indicates that the energy emitted from the hole is essentially blackbody radiation.

26.4

Radiation Intensity and Lambert’s Cosine Law

In Figure 26.6, we consider that dS1 is a diffuse emitting surface that emits radiation in all directions. We will use the intensity of radiation, designated by I , as the quantity that describes how this radiation varies with respect to the azimuthal angle d and the zenith φ

dS2

dφ

dS1

dψ ψ

FIGURE 26.6 Configuration and nomenclature for the analysis of intensity of radiation.

860

Introduction to Thermal and Fluid Engineering

angle d in this spherical coordinate system. When placed above the emitting surface, dS1 , the hemisphere shown will intercept all of the energy emitted by dS1 but this emitted energy will only be “seen” without distortion from a point directly above dS1 . At the point dS2 , which is displaced by an angle of from the normal to dS1 , the element dS1 appears as the projected area dS1 cos . We define the intensity of radiation I at some point in space due to the emission from area dS1 as the radiant energy emitted per unit time per unit solid angle subtended at dS1 per unit of emitting surface normal to the direction of a point in space from the emitting source. If dS2 is considered at an angle from the normal, the solid angle subtended by dS2 as viewed from dS1 will be d = dS2 /r 2 and the heat flow from dS1 to dS2 will be ˙ = I cos dS1 d = I cos dS1 dS2 dQ r2

(26.11)

We note that dS1 cos is the effective or projected area of dS1 as viewed from dS2 and that ˙ dQ dS1 cos d

I =

(W/m2 -sr)

(26.12)

The total emissive power will be a summation of the intensity over the entire hemispherical surface. In the case of Figure 26.6, the radiation intercepted by the element dS2 is integrated over the half-space represented by the hemisphere     dS2 E= I cos d d = I cos 2 d (26.13) r   With regard to Figure 26.6, an exercise in spherical coordinates will show that d =

(r sin )(d )(r d) r2

or d = sin d d so that, with this substituted for d in the integral in Equation 26.13, we obtain 



2

E=

/2

d 0

I cos sin d

0

First, we integrate with respect to  to obtain  E = 2I

/2

cos sin d

0

and then, when we integrate with respect to , the result is

/2  

1 2

E = 2I sin

2 0 or E = I

(26.14)

Equation 26.14 is Lambert’s law and shows that the total emissive power is equal to  times the radiation intensity.

Radiation Heat Transfer

861

ε

1.00 0.75 0.50 0.25

0

1

2

3

4

5

6

7

8

λ, μm

FIGURE 26.7 Emissivity for Example 26.2.

26.5

Monochromatic and Total Emissivity and Absorptivity

26.5.1 Emissivity If E  (T) is denoted as the hemispherical monochromatic emissive power of a nonblack surface, then the hemispherical monochromatic emissivity is defined as the ratio of E  (, T) to the hemispherical monochromatic emissive power of a black surface at the same temperature and wavelength

 (, T) =

E  (, T) E b (, T)

(26.15)

An even more global measure of the difference between a real surface and a black surface at the same temperature is the total hemispherical emissivity, (T)  ∞  ∞

 (, T) E b (, T)d E(T) 1 0

(T) = = E (, T)d = (26.16)  E b (T) T 4 0 T 4

Example 26.2 A diffuse emitter at 1800 K possesses a discontinuous hemispherical

monochromatic emissivity of 0.25 from 0 m to 2 m, 0.75 from 2 m to 6 m and 0.50 above 6 m. Determine (a) the total hemispherical emissivity and (b) the total emissive power.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) The surface is a diffuse emitter. (3) Radiation is the only mode of heat transfer. (a) The total hemispherical emissivity is given by Equation 26.16. Because the values of the monochromatic emissive power are constants within the three wavelength bands, we

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Introduction to Thermal and Fluid Engineering

may write 

1

=



1 T

2

E b (, T) 0

+

T 4



2 , T

3

E b (, T)

1 , T

+

T 4

∞

2 , T

E b (, T)

T 4

or

= 1 F0-1 T + 2 ( F0-2 T − F0-1 T ) + 3 ( F0-3 T − F0-2 T ) With 1 T = (2 -m)(1800 K) = 3600 m-K 2 T = (6 -m)(1800 K) = 10,800 m-K and 3 T = ∞, Table 26.1 gives F0-3600 = 0.4036,

F0-10,800 = 0.9283,

and

F0-∞ = 1.0000

Thus,

= 0.25(0.4036) + 0.75(0.9283 − 0.4036) + 0.50(1.0000 − 0.9283) = 0.1009 + 0.75(.5247) + 0.50(0.0717) = 0.1009 + 0.3935 + 0.0359 = 0.530 ⇐ (b) The total emissive power is E = T 4 = 0.530(5.670 × 10−8 W/m2 -K4 )(1800 K) 4 = 315.6 kW/m2 ⇐ 26.5.2 Absorptivity In dealing with the hemispherical monochromatic emissivity, we consider the hemispherical irradiation, G  , incident on a surface and define the hemispherical monochromatic absorptivity as the fraction of this incident irradiation that is absorbed, G a  (, T)  (, T) =

G a  (, T) G  (, T)

(26.17)

Here we note that the absorptivity is also a function of both the wavelength and the temperature but that the wavelength to be employed is the wavelength of the incident irradiation. The total hemispherical absorptivity will be  (T) =

G a (T) = G(T)

0

∞

G a  (, T)d G(T)

(26.18)

Radiation Heat Transfer

863

26.5.3 The Gray Surface or Gray Body A gray surface or gray body with temperature, T, is a surface or body whose hemispherical monochromatic emissivity is independent of wavelength

 (, T) =  (T)

(26.19)

This infers that, in addition to satisfying the requirement of Equation 26.19, the gray surface or body is an opaque diffuse emitter, absorber, and reflector.

26.6

Heat Flow between Blackbodies

26.6.1 The Shape Factor The Stefan-Boltzmann law applies to emission from a single surface. In this section, we consider the interchange of thermal energy between two surfaces such that not all of the radiation from one is intercepted by the other. Consider Figure 26.8, which shows two surfaces, S1 and S2 , separated by a nonabsorbing medium. The distance between the emitting surface, dS1 , and the receiving surface, dS2 , is designated by r . The rate of radiant energy interchange between dS1 and dS2 is ˙ 12 = I1 cos 1 dS1 d12 dQ where d12 is the solid angle subtended by dS2 with respect to dS1 . This is equal to the projected area of the receiving surface divided by the square of the distance between the surfaces dS2 cos 2 r2

d12 =

S2 dS2 φ2

r

φ1 S1 dS1

FIGURE 26.8 Geometric shape factor notation.

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Introduction to Thermal and Fluid Engineering

so that ˙ 12 = dQ

I1 cos 1 cos 2 dS1 dS2 r2

and because I1 = E 1 / ˙ 12 = E 1 dS1 dQ



cos 1 cos 2 dS2 r 2

 (26.20a)

and we consider the term within the parentheses as the fraction of the total emission from dS1 that is intercepted by dS2 . It can be shown in a similar manner that   ˙ 21 = E 2 dS2 cos 1 cos 2 dS1 dQ (26.20b) r 2 so that the net rate of radiant energy interchange between the differential elements can be obtained from the difference of Equations 26.20a and 26.20b   cos 1 cos 2 dS1 dS2 ˙ d Q = ( E1 − E2) r 2 or ˙ 12 = ( E 1 − E 2 ) Q

  S1

S2

cos 1 cos 2 dS1 dS2 r 2

(26.21)

The double integral may be written as S1 F12 , where F12 is called the shape factor (also referred to as the view factor, the configuration factor, or the arrangement factor) with respect to area S1 . The value of F12 may be obtained from   1 cos 1 cos 2 F12 = dS1 dS2 (26.22a) S1 S1 S2 r 2 and, in a similar fashion, we may define F21 as   1 cos 1 cos 2 F21 = dS2 dS1 S2 S2 S1 r 2

(26.22b)

and with the shape factor defined in this manner, the heat interchange will be ˙ 12 = F12 S1 ( E 1 − E 2 ) Q or in the case of heat exchange from body 2 to body 1 ˙ 21 = F21 S2 ( E 2 − E 1 ) Q In general, ˙ = Si Fi j E Q

(26.23)

S1 F12 = S2 F21

(26.24)

and the equality

is known as the reciprocity theorem.

Radiation Heat Transfer

865

0.9 Y/H = 20.0

0.8

2

X

Y

0.6 F1–2

10.0 6.0 4.0 3.0

1

H 0.7

2.0

0.5

1.5

0.4

1.0

0.3

0.8

0.2

0.6 0.4

0.1

0.2 0.1

0

1.0

2.0

3.0 X/H

4.0

5.0

6.0

FIGURE 26.9 Shape factor between directly opposed rectangles.

The determination of the shape factor by means of the evaluation of the double integral is tedious even for the most simple configurations. Fortunately, however, the literature contains many references to specific configurations of practical interest (Howell, 1982; Mahan, 2002; Modest, 2003; Siegel and Howell, 2001). Figures 26.9 through 26.11 provide graphs of the shape factor for three common arrangements and relationships for nine shape factors are listed in Section 26.6.2.

Example 26.3 Determine the shape factor between a small area, dS1 and a large, parallel area, S2 of diameter, D. The areas are aligned on their centerlines and are R units apart.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) Both surfaces are diffuse. (3) Radiation is the only mode of heat transfer. (4) S2  S1 . The configuration is displayed in Figure 26.12 where we note that the area dS2 is in the shape of a ring with an area dS2 = x dxd and that 1 = 2 = ,

r = ( R2 + x 2 ) 1/2 ,

and

cos =

R ( R2 + x 2 ) 1/2

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Introduction to Thermal and Fluid Engineering

2

Z

1

X

Y 0.5 Y/X = 0.05 0.4

0.1 0.2

0.3 F12

0.4

0.6

0.8

1.0

0.2

1.5 2.0 3.0

0.1

0.1

0.2

0.3

0.4 0.5 0.6

0.8 1.0 Z/X

2.0

3.0

4.0 6.0 10.0

4.0 0.5 0.6

20.0 0.8 10.0

FIGURE 26.10 Shape factor between perpendicular rectangles with a common edge.

1.0

1.0 L

0.9

0.9 r2

0.8 Y=∞ 0.7

Y2 = 5

r1

F2–1 0.8

X = r1/r2 Y = L/r2

0.7

Y2 = ∞

F2–2

0.6

0.6 2

0.5

2 0.5

0.4 0.3

0.1

0.4

0.3 0.2

1

0.2

0.5

1

0.3

0.1 0.5

0.05

0.2

0.3 0.2

0.1

0.1 0.05 0

0.1

0.2

0.3

0.4

0.5 X

FIGURE 26.11 Shape factor between infinitely long concentric cylinders.

0.6

0.7

0.8

0.9

1.0

F2–1

F22

5

Radiation Heat Transfer

867 d x

dS1

dx

φ2

R r = (R2 + x2)1/2

φ1 dS2 FIGURE 26.12 Radiation from a small area element to a disk for Example 26.3.

The shape factor will be given by Equation 26.22a F12 =

1 S1

  S1

S2

cos 1 cos 2 dS1 dS2 r 2

(26.22a)

and with the foregoing substitutions  F12 = S2

cos2 dS2 = r 2

 0

d/2

2R2 xd x ( R2 + x 2 ) 1/2

or F12 =

d2 ⇐ 4R2 + d 2

26.6.2 A Catalog of Simple Shape Factors in Two Dimensions A tabulation of the relationships for the computation for nine two-dimensional shape factors extracted from Howell (1982) are given by Equations 26.25 to 26.33. By two dimensional, we mean that the dimension into the plane of the paper is infinite. The configurations for the nine cases are illustrated in Equations 26.24 through 26.33.

1. Two infinitely long plates of equal width joined at an edge F12 = F21 = 1 − sin

 2

(Figure 26.13)

(26.25)

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Introduction to Thermal and Fluid Engineering

W2 S2 S1

α W1 FIGURE 26.13

2. Two infinitely long plates of different widths, W1 and W2 , joined at a common edge at an angle of 90◦

F12

⎫ ⎧    1/2 ⎬ 1⎨ W1 W1 2 = 1+ − 1+ ⎭ 2⎩ W2 W2

(Figure 26.14)

(26.26)

S2 W1 90° S1

W2 FIGURE 26.14

3. Triangular enclosure formed from three infinitely long plates of widths W1 , W2 , and W3 F12 =

W1 + W2 − W3 2W1

(Figure 26.15)

W2

W3 S2 S3 S1

W1 FIGURE 26.15

(26.27)

Radiation Heat Transfer

869

4. Disk of radius R and infinitesimal parallel disk positioned H units away on the disk centerline F12 =

R2 H 2 + R2

(Figure 26.16)

(26.28)

dS2

H S1

R1 FIGURE 26.16

5. Two parallel disks of radii R1 and R2 positioned H units apart on the same centerline. With x1 = R1 /H,

F12

x2 = R2 /H,

and

⎫ ⎧   2 1/2 ⎬ 1⎨ x 2 = X − X2 − 4 ⎭ 2⎩ x1

R2

S2

H

S1

R1 FIGURE 26.17

X=1+

1 + x22 x12

(Figure 26.17)

(26.29)

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Introduction to Thermal and Fluid Engineering

6. Infinitely long cylinder parallel to infinitely long plate with finite width, W F12 =

  1 L1 L2 arctan − arctan 2 H H

(Figure 26.18)

(26.30)

R

H

L2 L1 FIGURE 26.18

7. Two parallel and infinite cylinders of radius R with a clear spacing between them of L. With X = 1 + L/2R and Y = ( X2 − 1) 1/2 F12 = F21 =

  1 1 Y + arcsin − X  X

(Figure 26.19)

(26.31)

R

L

R FIGURE 26.19

8. Row of equidistant infinitely long cylinders of diameter D with centerline spacing of W parallel to an infinite plane. With x = D/W 

F12 = 1 − (1 − x )

2 1/2

1 − x2 + x arctan x2

1/2 (Figure 26.20)

(26.32)

Radiation Heat Transfer

871 S2

W D

S1 FIGURE 26.20

9. Sphere of radius R1 positioned H units on the same centerline from a disc of radius R2 . With X = R2 /H F12 =

 1 1 − (1 + X2 ) −1/2 2

(Figure 26.21)

(26.33)

S1 R1

Sphere H S2

disc

R2 FIGURE 26.21

26.6.3 A Catalog of Simple Shape Factors in Three Dimensions The relationships for three-dimensional shape factors with no restrictions placed on the dimension into the plane of the paper are provided in Equations 26.34 through 26.36.

1. Directly opposed rectangles of side dimensions X and Y located H units apart (Figure 26.9) with x = X/H and y = Y/H F12 =

2 [ϒ1 + ϒ2 + ϒ3 − ϒ4 ] xy

(26.34)

872

Introduction to Thermal and Fluid Engineering where  ϒ1 = ln

(1 + x 2 )(1 + y2 ) 1 + x 2 + y2

1/2

ϒ2 = x(1 + y2 ) 1/2 arctan

x (1 + y2 ) 1/2

ϒ3 = y(1 + x 2 ) 1/2 arctan

y (1 + x 2 ) 1/2

ϒ4 = x arctan x + y arctan y 2. Perpendicular rectangles with a common edge (Figure 26.10) with x = X/Z and y = Y/Z   1 1 F12 = (ϒ1 − ϒ2 + ln (ϒ3 ϒ4 ϒ5 ) (26.35) y 4 where ϒ1 = y arctan

1 1 + x arctan y x

ϒ2 = (x 2 + y2 ) 1/2 arctan ϒ3 =

(x 2

1 + y2 ) 1/2

(1 + x 2 )(1 + y2 ) 1 + x 2 + y2 

 y2



x2

y2 (1 + x 2 + y2 ) ϒ4 = (1 + y2 )(x 2 + y2 ) x 2 (1 + x 2 + y2 ) ϒ5 = (1 + x 2 )(x 2 + y2 )

3. Coaxial cylinders of finite length (Figure 26.11) with x = r2 /r1 , y = L/r1 , A = y2 + x 2 − 1, and B = y2 − x 2 + 1   1 1 1 F21 = − ϒ1 − (26.36a) (ϒ2 + ϒ3 ) x x 2y where ϒ1 = arccos

B A

ϒ2 = ( A2 + 4A − 4x 2 + 4) 1/2 arccos ϒ3 = B arcsin

B Ax

1 A − x 2

and F22 = 1 −

1 y 2 + ϒ4 − (ϒ5 − ϒ6 ) x x 2x

(26.36b)

Radiation Heat Transfer

873

where ϒ4 = arctan

2(x 2 − 1) 1/2 y

(4x 2 + y2 ) 1/2 4(x 2 − 1) + ( y2 /x 2 )(x 2 − 2) arcsin y y2 + 4(x 2 − 1)   x 2 − 2  (4x 2 + y2 ) 1/2 ϒ6 = arcsin + − 1 x2 2 y

ϒ5 =

26.6.4 Properties of the Shape Factor While the shape factor can be determined (often with great difficulty) from the relationships of Equation 26.22a,b, it is possible to deduce their value from one or more known factors that pertain to a related geometry. The use of this procedure is referred to as making use of shape factor algebra and is based on the reciprocity, additivity, and enclosure properties. 26.6.4.1 The Reciprocity Property A modification of Equation 26.24 introduced in Section 26.6.1 S1 F12 = S2 F21

(26.24)

is known as the reciprocity theorem and is the reciprocity property. It shows that, in a twobody system involving diffuse black surfaces, the heat emitted by the hot surface is equal to the heat absorbed by the cooler surface. 26.6.4.2 The Additivity Property The additivity property considers the radiation interchange between a surface and another surface that may be subdivided into n subsurfaces F12 =

n 

F12,i

(26.37)

i=1

If, for example, surface 2 is subdivided into three subsurfaces, S2a , S2b , and S2c , the additivity property tells us that F12 = F12a + F12b + F12c which shows that the shape factor F12 is the shape factor between surface 1 and all of the pieces of surface 2. Equation 26.37 describes the additivity property. 26.6.4.3 The Enclosure Property Suppose that a surface S1 , which may be concave so that it “sees” itself is completely surrounded by an enclosure containing n surfaces. Then the blackbody emission from surface 1, E b1 , will be intercepted by each surface, Si and will be a fraction of E bi . Hence, E b1 S1 = F11 S1 E b1 + F12 S1 E b1 + F13 S1 E b1 + · · · and after we divide by the product E b1 S1 we have 1 = F11 + F12 + F13 + · · ·

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Introduction to Thermal and Fluid Engineering

or n 

Fi j = 1

(i = 1, 2, 3, . . . , n)

(26.38)

j=1

Equation 26.38 describes the enclosure property and, in it, we note that the subscript, i, refers to the emitting surface and the subscript, j, accounts for all of the other surfaces that contribute to the enclosure. We reemphasize that if none of the surfaces “see” themselves, then F11 , F22 , F33 , . . . , Fnn will not be present. A simple application of the reciprocity and enclosure properties is now provided in Example 26.4.

Example 26.4 Consider (a) a pair of infinite rectangular plates, S1 and S2 and (b) an infinite cylinder, S1 , within another infinite cylinder, S2 . In both cases, displayed in Figure 26.22, the end effects are negligible. Determine F12 , F21 , F11 , and F22 .

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) All surfaces are diffuse. (3) Radiation is the only mode of heat transfer. (4) The plates and cylinders are of infinite extent. (a) Because the plates are of infinite extent with S1 = S2 and since the end effects are negligible, by reciprocity F12 = F21 = 1 ⇐ Then, because the plates don’t “see” themselves F11 = F22 = 0 ⇐ (b) All of the radiation from S1 is intercepted by S2 . Hence, F12 = 1 ⇐ S1 S2

S1

S2 (a)

(b)

FIGURE 26.22 Configurations for Example 26.4: (a) an infinite pair of rectangular plates and (b) a pair of infinite cylinders.

Radiation Heat Transfer

875

and by reciprocity S1 F12 = S2 F21 so that F21 =

S1 S1 F12 = ⇐ S2 S2

Because surface 1 does not see itself F11 = 0 ⇐ and by the enclosure property F22 = 1 − F21 = 1 −

S1 ⇐ S2

A straightforward application of the additivity property is provided in Example 26.5.

Example 26.5 Determine the shape factor F14 for the configuration of perpendicular rectangles indicated in Figure 26.23.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) All surfaces are diffuse. With S5 = S1 + S2

S6 = S3 + S4

and

we have S5 F54 = S1 F14 + S2 F24 or S1 F14 = S5 F54 − S2 F24

S1

S2

S3

S4

FIGURE 26.23 Configuration for Example 26.5.

(a)

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Introduction to Thermal and Fluid Engineering

Then S5 F56 = S5 F54 + S5 F53 or S5 F54 = S5 F56 − S5 F53

(b)

and S2 F26 = S2 F24 + S2 F23 or S2 F24 = S2 F26 − S2 F23

(c)

When we put (b) and (c) into (a), we obtain the result sought, which is in terms of shape factors that can be obtained from a graph, Equation 26.35 or software F14 =

1 [S5 ( F56 − F53 ) − S2 ( F26 − S2 F23 )] ⇐ S1

26.6.5 The Symmetry Property If two surfaces, j and k, possess the same surface area and are symmetric about the surface, i, then Fij = Fik .

26.7

Heat Flow by Radiation between Two Bodies

26.7.1 Diffuse Blackbodies For two diffuse blackbodies having surfaces, S1 and S2 , the radiant heat exchange will be   E b1 − E b2 =  T14 − T24 (W/m2 ) where T1 > T2 . This heat exchange may be multiplied by the surface area of either body with appropriate cognizance taken of the shape factor   ˙ 12 = S1 F12 T14 − T24 = S2 F21 (T14 − T24 ) (W) Q (26.39a) Moreover, if the surfaces are in full view of one another so that F12 = F21 = 1 ˙ 12 = S1 (T14 − T24 ) Q

(W)

(26.39b)

26.7.2 Opaque Gray Bodies If two gray surfaces are in full view of one another such as in the case of infinite parallel planes where F12 = F21 = 1, Equation 26.39b will not hold because the surfaces are not black. If the surfaces are opaque and gray, then =1−

and

=

Radiation Heat Transfer

877

Eb1

α1 ε2 Eb2

α1 ρ2 ε1 Eb1

ε1 E

b1

ρ1 ε 2E b

ρ1 ρ

ρ12ρ

2

Eb2

α2 ε1 Eb1 2

ρ2

ε 1 E b1

α2ρ1 ε2 Eb2

2 ε1

α1 ρ2ρ1 ε2 Eb2

ε 2 E b2

Eb

1

ρ 2ρ 1

ε 2 E b2

α2 ρ1ρ2 ε1 Eb1 ε2 E

b2

ε 1 E b1 2 ρ 2ρ 1

α1 ρ22ρ1 ε1 Eb1

α2 ρ21ρ2 ε2 Eb2

FIGURE 26.24 Reflection and absorption characteristics associated with two infinite parallel gray surfaces.

If surface 1 emits 1 E b1 , surface 2 will absorb the fraction 2 1 E b1 and reflect 2 1 E b2 back toward surface 1. Further reflections will occur and they are indicated in Figure 26.24 where a similar picture is shown for surface 2. The net heat transferred per unit of surface 1 to surface 2 will be the original transmission

1 E b1 minus the fraction of 1 E b1 and of 2 E b2 , which is ultimately absorbed by surface 1 after successive reflections. Hence, we have with 1 = 1−1 = 1− 1 and 2 = 1−2 = 1− 2 ˙ 12 = S1 1 E b1 [1 − 1 (1 − 2 ) − 1 (1 − 1 )(1 − 2 ) 2 + · · · ] Q −S2 2 E b2 [ 1 + 1 (1 − 1 )(1 − 2 ) + 1 ((1 − 1 ) 2 (1 − 2 ) 2 + · · · ] and, if S1 = S2 = S, the two infinite series reduce such that   ˙ 12 = S( E b1 − E b2 ) 1 + 1 − 1 Q

1

2

(26.40)

It is convenient to define a factor F12 (apparently first introduced by Hottel, 1954) to handle the fact that the surfaces are gray   ˙ 12 = SF12 T14 − T24 Q (26.41) where F12 can be considered as an emissivity factor. This factor is to be used for opaque gray configurations with surfaces in full view of one another in the same way as the shape factor is used for diffuse black surfaces. Indeed, for infinite parallel planes F12 =

1 1 1 + −1

1

2

(26.42)

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Introduction to Thermal and Fluid Engineering

and it can be shown that for a small body having surface, S1 , within a large enclosure, S2 F12 = 1

(26.43)

and for concentric cylinders or spheres, S1 surrounded by S2 F12 =

1 S1 1 1+ (1 − 2 ) S2 2

(26.44)

In this section, we have considered diffuse black surfaces (ideal emitters and absorbers) in geometries where not all of the radiant energy emitted by the source is intercepted by the receiver and gray opaque surfaces in full view of one another. Both of these considerations have pertained to a single source and a single receiver. The next two sections lead to an electrothermal analog due to Oppenheim (1956), which leads to expeditious calculations for multiple radiating and reradiating surfaces.

Example 26.6 A very thin radiation shield is to be placed between two opaque gray parallel planes of infinite extent as indicated in Figure 26.25. The shield has two surfaces denoted by the subscripts  and r (for “left” and “right”). Determine the temperature of the shield if conduction through the shield is to be neglected and T1 = 1000 K

1 = 0.800

 = 0.625

T2 = 400 K

2 = 0.750

r = 0.400

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) All surfaces are opaque, gray, and diffuse. (3) Radiation is the only mode of heat transfer. (4) End effects are negligible. In the steady state, the heat gained by the shield   ˙ 1 = SF1 T14 − Ts4 Q T1 = 1000 K ε1 = 0.800 εℓ = 0.625

FIGURE 26.25 Configuration for Example 26.6.

T2 = 400 K ε2 = 0.750 εr = 0.400

Radiation Heat Transfer

879

is equal to the heat lost by the shield   ˙ r 2 = SFr 2 Ts4 − T24 Q where Ts is the shield temperature. A little algebra provides     F1 T14 − Ts4 = Fr 2 Ts4 − T24   Fr 2 Fr 2 4 1+ Ts4 = T14 + T F1 F1 2 or ⎡

⎤ Fr 2 4 1/4 4 T + T ⎢ 1 F1 2 ⎥ ⎥ Ts = ⎢ ⎣ Fr 2 ⎦ 1+ F1 Equation 26.40 can then be employed to give F1 = 0.541,

Fr 2 = 0.353,

and Fr 2 /F1 = 0.653

which leads to a shield temperature of 885.6 K ⇐

26.8

Radiosity and Irradiation

We have seen that the emissive power, whether total or monochromatic, refers only to the original emission from a surface. It does not include any additional components that are due to reflection or multiple reflections of any of the incident radiation. Radiosity, denoted by J , is the term used to indicate all of the radiation leaving a surface (per unit time and per unit surface area). The total radiosity, J , may be obtained from the monochromatic radiosity, J  , via  J =



J  d

0

Moreover, we have seen in Section 26.3.1 that irradiation is the term used to denote the rate at which thermal radiation (again per unit time and per unit surface area) is incident upon a surface. This incident radiation, designated by G, may be the result of emissions and reflections from other surfaces. The total irradiation may be obtained from the monochromatic radiation, G  , from  G= 0



G  d

Both radiosity and irradiation are used in the next section, which employs a network method to discuss radiation within enclosures.

880

26.9

Introduction to Thermal and Fluid Engineering

Radiation within Enclosures by a Network Method

For surfaces within an enclosure that obey Lambert’s cosine law and reflect diffusively, a “network method” is quite useful for the evaluation of the multiple heat flow paths from surface to surface. We begin with the radiosity of the i surface J i = i G i + i E bi

(26.45)

where G i is the irradiation, i and i are the reflectivity and emissivity, and E bi is the blackbody emissive power. If both surface temperature and irradiation over the surface are uniform, then the net rate of heat loss or gain will be a function of the difference between the radiosity and the irradiation ˙ i = Si ( J i − G i ) Q

(26.46)

Moreover, if the surfaces are gray, we have the important limiting assumptions

i = i

and i = 1 − i

Equations 26.45 and 26.46 may be combined after we first observe from Equation 26.45 that Gi = and hence,

1 ( J i − i E bi ) i

  ˙ i = Si J i − 1 ( J i − i E bi ) Q i

or ˙ i = Si [ i E bi + (i − 1) J i ] = Si i ( E bi − J i ) Q i 1 − i

(26.47)

When we recast Equation 26.47 into the form ˙ i = E bi − J i Q 1 − i Si i

(26.48)

we see that an analogy may be drawn that shows E bi − J i as a difference of potential, and the fraction (1 − i )/Si i as a radiation thermal resistance. This resistance governs the flow of ˙ i , between a node point representing the blackbody emission, E bi = Ti4 radiant energy, Q and another node point representing the radiosity, J i . The direct radiative interchange between surface i and surface j is governed by the difference between the two radiosities, J i and J j , the shape factor, Fij , and the surface, Si , according to ˙ ij = J i − J j Q 1/Si Fij The reciprocity relationship, Equation 26.19, will allow this to be written as ˙ ij = J i − J j = J i − J j Q 1/Si Fij 1/S j Fji

(26.49)

Radiation Heat Transfer

881

S2

S3

S1 (a) Eb2

1–ε2 S2 ε2

1–ε2 S3 ε3

1 S2F23 J2

J3

1 S1F12

1 S1F13

Eb3

J1 1– ε1 S1 ε1 Eb1 (b) FIGURE 26.26 (a) An arrangement of three radiating surfaces and (b) the electrothermal analog network.

Here, the radiative thermal surface-to-surface resistance is 1/Si Fij = 1/S j Fji . This resistance governs the flow of heat between two node points, each represented by a radiosity. Equations 26.48 and 26.49 form the basis for the electrical analog network, which was first proposed by Oppenheim (1956). An enclosure with three surfaces is shown in Figure 26.26a and the electrothermal analog network is provided in Figure 26.26b. The nodes marked E b1 , E b2 , and E b3 represent heat sources and each of these heat sources may be related to a temperature, E bi = Ti4 . The resistances (1 − i )/Si e i “transfer” the E bi ’s to the surface radiosities, J i , thereby accounting for the fact that the surfaces are not black but gray. The resistances, 1/Si Fij = 1/S j Fji , link the radiosities and account for the actual radiative interchange between surface i and surface j. Example 26.7 shows the set-up of the actual circuit. It should be kept in mind that the evaluation of the radiosities is only rarely required and that the object of the game is the determination of the surface temperatures.

Example 26.7 Use the network method to determine the temperature of the shield in Example 26.6.

Solution Assumptions and Specifications (1) Steady-state conditions exist. (2) All surfaces are opaque, gray, and diffuse. (3) Radiation is the only mode of heat transfer. (4) End effects are negligible.

882

Introduction to Thermal and Fluid Engineering Eb1 = σT14 R1 =

1–ε1 ε1

J1 R2 =

1 F1s

J2

R3 =

Eb1–Eb2

+

1–εℓ εℓ

Ebs

– R4 =

1–εr εr

J3 R5 =

1 Fs2

J4 R6 =

1–ε2 ε2

Eb2 = σT24 FIGURE 26.27 The electrothermal analog network for Example 26.7.

The analog circuit is displayed in Figure 26.27 where we note the designators for the various resistances. With the same data as in Example 26.6, T1 = 1000 K

1 = 0.800

 = 0.625

T2 = 400 K

2 = 0.750

r = 0.400

we have for S = 1 m2 and both of the walls in full view of the shield R1 =

1 − 1 1 − 0.800 = = 0.2500 m−2

1 0.800 R2 =

R3 =

1 1 = = 1.0000 m−2 F1s 1

1 −  1 − 0.625 = = 0.6000 m−2

 0.625

Radiation Heat Transfer

883 R4 =

1 − r 1 − 0.400 = = 1.5000 m−2

r 0.400 R5 =

1 1 = = 1.0000 m−2 Fs2 1

and R6 =

1 − 2 1 − 0.750 = = 0.3333 m−2

2 0.750

and the sum of these six resistances is 

R = 4.6833 m−2

The heat flow from wall 1 to wall 2 will be

   T14 − T24 E b1 − E b2 ˙   Q12 = = R R

or 2 4 4 −8 4 ˙ 12 = (5.67 × 10 W/m -K )[(1000 K) − (400 K )] Q 4.6833 m−2

= (1.211 × 10−8 W/K4 )[1.000 × 1012 K4 − 2,560 × 1010 K4 ] = (1.211 × 10−8 W/K4 )(9.744 × 1011 K4 ) = 11.797 kW We can obtain the shield temperature from the temperature drop through the first three resistances ˙ 12 ( R1 + R2 + R3 ) E b1 − E bs = Q = (11.797 kW)(0.250 m−2 + 1.000 m−2 + 0.600 m−2 ) = (11.797 kW)(1.850 m−2 ) = 21.83 kW/m2 Thus, E bs = E b1 − 21.83 kW/m2 = T14 − 21.83 kW/m2 = (5.67 × 10−8 W/m2 -K4 )(1000 K) 4 − 21.83 kW/m2 = 56.70 kW/m2 − 21.83 kW/m2 = 34.87 kW/m2 Then Ts4 =

34.87 kW/m2 34.87 kW/m2 = = 6.151 × 1011 K4  5.67 × 10−8 W/m2 -K4

884

Introduction to Thermal and Fluid Engineering

and Ts = 885.6 K ⇐ This value checks the value obtained in Example 26.6 and we note that for those of the electrical persuasion, the voltage divider will work as long as due cognizance is taken of the fourth-power dependency of the emissive powers.

26.10

Summary

Thermal radiation is but one form of electromagnetic radiation. The narrow band of radiation between 0.36 <  < 0.76 m is known as the visible region, whereas the band between 0.1 m < 100 m is generally considered as thermal radiation; and this range includes the infrared and a portion of the ultraviolet components of the electromagnetic spectrum. A blackbody is an ideal surface that absorbs all and reflects none of the radiation that falls upon it. The monochromatic emissive power, which is a function of both wavelength and temperature, is given by Planck’s law E b (, T) =

2C1 c 2 5 cC  (e 1 /C2 T −

1)

(W/m2 )

(26.2)

where C1 = 6.624 × 10−27 erg and C2 = 1.380 × 10−16 erg/K The wavelength of maximum monochromatic emissive power at a particular temperature is given by Wien’s displacement law T = 2897.8 -K

(26.3)

and the quantity of blackbody radiation at a particular temperature is given by the StefanBoltzmann law E b = T 4

(26.4)

The fraction of emission falling between wavelengths of 1 and 2 is given by F1 -2 = F0-2 T − F0-1 T

(26.6)

with values tabulated as a function of T in Table 26.1. Three system properties that concern radiation and pertain to a surface are the absorptivity, , the transmissivity, , and the reflectivity, . These have a magnitude less than or equal to unity and are related by + +=1

(26.7)

Radiation Heat Transfer

885

Kirchhoff’s law states that for n small blackbodies in thermal equilibrium in a blackbody enclosure E1 E2 En Eb = = + ··· + 1 2 n The emissivity of a surface is related to the blackbody emissive power and is defined by

=

E Eb

and for a small body emitting E 1 in a blackbody enclosure at the same temperature 1 = 1 Lambert’s cosine law for the intensity of radiation is E = I

(26.16)

A gray body having a gray surface is a surface whose emissivity is independent of wavelength. The shape factor (also referred to as the view factor, the configuration factor, or the arrangement factor) between two surfaces accounts for the fact that not all of the radiation emitted by one surface is intercepted by the other. Several shape factors are provided in Sections 26.6.1 and 26.6.2. Shape factors have properties that allow an analyst to employ shape factor algebra. These properties are the reciprocity property S1 F12 = S2 F21

(26.19)

the additive property F12 = F12a + F12b + F12c where surface 2 is divided into n subsurfaces, the enclosure property n 

Fi j = 1

(i = 1, 2, 3, . . ., n)

(26.38)

j=1

and the symmetry property, which states that if two surfaces j and k possess the same area and are symmetric about a surface, i, then Fi y = Fik . Radiosity, denoted by J, is defined as the total radiation leaving a surface and irradiation, denoted by, G, is defined as the total radiation incident on a surface. An electrothermal analog method may be employed for situations where the surfaces are not in full view of one another and where the surfaces are not blackbodies. The analog is based on surface blackbody node points represented by E b,i and their associated radiosities, J i . The connection between E b,i and J i is governed by a resistance, R = (1 − 1 )/Si i . The radiosities, J i and J j , are connected by a resistance, R = 1/Si Fij = 1/S j Fji .

26.11

Problems

Emissive Power and Spectral Characteristics 26.1: Determine the monochromatic emissive power for a black surface at a wavelength of 10 m at temperatures of (a) 200 K, (b) 500 K, and (c) 1000 K. 26.2: In an experiment, the monochromatic emissive power of a blackbody is measured to be 5010 W/m2 -m-K4 at a wavelength of 1.5 m. Determine (a) the temperature of

886

Introduction to Thermal and Fluid Engineering the blackbody, (b) the wavelength at which the maximum monochromatic emissive power occurs, and (c) the maximum monochromatic emissive power.

26.3: Determine the peak wavelengths and the peak monochromatic emissive power of black surfaces at (a) 400 K, (b) 1200 K, (c) 3000 K, and (d) 3500 K. 26.4: Determine the radiation emitted, in W/m2 between 4 and m by black surfaces at (a) 400 K, (b) 800 K, and (c) 1200 K. 26.5: The fraction of the radiation emitted between 2 m and a higher but unknown wavelength, , is 0.25 for a black surface at 800 K. Determine the unknown wavelength. 26.6: Determine the total emissive power of a blackbody at 600◦ C and 1200◦ C. 26.7: The so-called solar constant measured at the fringe of the earth’s atmosphere located 93 × 106 miles from the center of the sun is 1394 W/m2 . Determine its value at (a) Venus (mean orbital radius of 36 × 106 miles) and (b) Pluto (mean orbital radius of 3670 × 106 miles). 26.8: Suppose that the filament of an ordinary 100-W lightbulb is at 2910 K and that the bulb can be approximated as a blackbody. Determine (a) the wavelength of maximum emission and (b) the fraction of emission in the visible region of the spectrum. 26.9: For a blackbody maintained at 227◦ C, determine (a) the total emissive power, (b) the wavelength for maximum emissive power, (c) the magnitude of the maximum emissive power, and (d) the fraction of the emission between wavelengths of 1.25 and 4.25 m. 26.10: Rework Problem 26.9 for a blackbody temperature of 1027◦ C. 26.11: The radiation emitted by a blackbody at 620◦ C is 5408 W/m2 between wavelengths of  and  = ∞. Determine (a) the wavelength  and (b) the radiation emitted between  = 0 and  in part (a). 26.12: The inside surface of a large spherical enclosure is isothermal and maintained at 1200 K. Determine (a) the radiant heat flux, in W/m2 , from a small opening on the enclosure, (b) the fraction of this heat flux that is confined between 2 and 6 m, and (c) the maximum monochromatic emissive power associated with the emission. 26.13: A small circular hole is to be drilled in the surface of a large, hollow, spherical enclosure maintained at 2000◦ C. It is required that 100 W emerge from the hole. Determine (a) the hole diameter, (b) the number of watts emitted in the visible range from 0.6 to 0.7 m, (c) the ultraviolet range between 0.0 and 0.4 m, and (d) the infrared range between 0.7 and 100 m. 26.14: A 100-cm-diameter thin disk is maintained at a uniform temperature of 427◦ C. Both surfaces of the disk behave as black surfaces. Determine the total radiant power, in watts, emitted from the disk in the wavelength bands (a) 2–4 m, (b) 6–8 m, and (c) 10–20 m. 26.15: Consider Planck’s law E b =

2C1 c 2 5 [e cC1 /C2 T − 1]

where C1 and C2 are constants and c is the speed of light. Show that for very large values of T, E b may be approximated by E b =

2cC1 cT C2 4

Radiation Heat Transfer

887

and for very small values of T the approximation will be E b =

2c 2 C1 5 e cC2 /T

26.16: A substantial cavity with a small opening that is 0.0025 m2 in area emits 8 W. Determine the wall temperature of the cavity. 26.17: Determine the wavelength of maximum emission for (a) the sun with an assumed temperature of 5790 K, (b) a lightbulb filament at 2910 K, (c) a surface at 1550 K, and (d) the human skin at 308 K. 26.18: For a blackbody surface at 1660 K, determine the fraction of the emission that occurs from (a) 0–1.25 m, (b) 1.25–3.25 m, (c) 3.25–6.50 m, and (d) 6.50–2.5 m. 26.19: The monochromatic emissive power of a surface varies with wavelength as = 0.125 for 0–0.625 m, = 0.515 for 0.625–5.15 m, = 0.700 for 5.15–12.5 m, and = 0.875 for 12.5–∞ m. Determine the total emissivity of the surface when the temperature is (a) 900 K, (b) 1800 K, and (c) 2700 K. 26.20: Assuming the sun to be a blackbody emitter at 5800 K, determine the fraction of the sun’s emission that falls in the visible region of the spectrum. 26.21: A furnace that has black interior walls maintained at 1227◦ C contains a peephole with a diameter of 10 cm. The glass in the peephole has a transmissivity of 0.78 between 0 and 3.2 m and 0.08 between 3.2 m and ∞. Determine the heat lost through the peephole. 26.22: Determine the temperature required for a blackbody emitter that will allow 40% of the energy emitted to fall between and 12 m. Radiation Properties and Kirchhoff’s Law 26.23: A horizontal opaque rectangular plate that is 15 cm × 20 cm receives radiation from its surroundings at the rate of 2000 W. Only one side of the plate is radiatively active and, the amount of radiation absorbed by the plate is 500 W. Determine (a) the irradiation, G in W/m2 , (b) the emissivity, , and (c) the reflectivity, . 26.24: A hemispherical cavity of radius 2 cm is machined into a block. The radiant heat flux leaving the cavity is 35,000 W/m2 , and the cavity has an emissivity of 0.70. Determine the inside surface temperature of the cavity. 26.25: An opaque circular plate, 8 cm in diameter is irradiated by 10 W on its top surface having a reflectivity of 0.22. The top surface is at a uniform temperature of 120◦ C. Determine (a) the radiant heat flux in W/m2 leaving the top surface and (b) the emissivity of the top surface. 26.26: In Problem 26.25, suppose that the top surface loses heat by convection to the surroundings at 20◦ C, which provides a heat transfer coefficient 0f 65 W/m2 -K. Determine the radiant heat flux in W/m2 . 26.27: A transparent plate is irradiated with 800 W of radiant energy. The amount of energy absorbed is 200 W. The reflectivity of the plate is 50% of its absorptivity. Determine the transmissivity of the plate. 26.28: The total radiant energy incident on a surface is 2400 W. The surface absorbs 825 W/m2 and reflects 600 W/m2 . Determine the transmissivity. 26.29: A set of reflectivity measurements for a solid surface at 800◦ C shows  = 0.12(0 <  < 3.2 m),  = 0.335(3.2 <  < 6.4 m),  = 0.615(6.4 < 8.9 m), and  = 0.98(8.9 <  < ∞ m). Determine the total emissive power.

888

Introduction to Thermal and Fluid Engineering

26.30: The reflectivity and transmissivity of a gray diffuse surface are respectively one half and one third of its absorptivity. Determine the absorptivity of the surface. 26.31: A triangular opaque horizontal plate having an emissivity of 0.72 is maintained at a temperature of 700◦ C. Each side of the plate measures 15 cm. The irradiation on the plate is 5200 W/m2 . The plate also loses heat to the surrounding air at 300 K through a heat transfer coefficient of 35 W/m2 -K. Determine (a) the reflected radiation, (b) the total energy leaving the plate, and (c) the net heat transfer from the plate. 26.32: An inclined plate receives 1100 W/m2 of radiant energy on its front surface. The back side of the plate is insulated. The plate has an absorptivity of 0.72 and an emissivity of 0.32. Determine (a) the steady temperature of the plate and (b) whether the plate can be considered a gray surface. 26.33: Reconsider the plate in Problem 26.32 under the following conditions: The plate also dissipates to the surroundings at 300 K by convection through a heat transfer coefficient of 27 W/m2 -K. Determine the steady-state temperature of the plate. 26.34: An opaque surface is initially at a uniform temperature of 500 K. The surface, having an absorptivity of 0.40 and an emissivity of 0.60, is suddenly exposed to a radiant energy flux of 12,000 W/m2 . Determine (a) whether the temperature increases or decreases with time and (b) how the picture changes, if at all, if the radiant energy flux is increased to 15,000 W/m2 . Radiation Properties and Lambert’s Cosine Law 26.35: Two small surfaces with areas S1 = 0.01 m2 and S2 = 0.02 m2 are oriented as shown in Figure P26.35. The surface S1 acts as a black surface at 800 K. Determine the rate at which radiant energy from S1 is intercepted by S2 .

S2 = 0.02 m2

r=

0.

60

m

45°

45°

S1 = 0.01 m2 T1 = 800 K ε1 = 0.78 FIGURE P26.35

26.36: Determine the monochromatic intensity of radiation, Ib , in W/m2 -sr- m at 5 m for a black surface at (a) 1000 K, (b) 800 K, and (c) 500 K. 26.37: Consider three small surfaces with areas, S1 = 0.01 m2 , S2 = 0.015 m2 , and S3 = 0.02 m2 oriented as shown in Figure P26.37. The emissive power of S1 is 29,600 W/m2 and the emission is diffuse. Determine the rate at which radiant energy from S1 is intercepted by S2 and S3 .

Radiation Heat Transfer

889 S3

r 31

S2

60°

m .40 =0

25°

30°

.80

r 21

=0

m

S1 FIGURE P26.37

26.38: Determine the maximum monochromatic intensity of radiation Ib for a black surface at 1000, 800, and 500 K. Use the results of Problem 26.36 to determine the ratio of monochromatic intensity of radiation to the maximum monochromatic intensity of radiation for each temperature. 26.39: Consider Figure P26.39 and observe that a small surface, S1 = 0.025 m2 , is maintained at 750 K and emits as a black surface. A second surface, S2 = 0.03 m2 , is to be oriented such that it captures 20% of the emission from S1 . Determine the angle of orientation, .

θ

S2

r=1m 60°

S1 = 0.025 m2 FIGURE P26.39

Monochromatic and Total Emissivity and Absorptivity 26.40: A diffuse emitter at 1500 K has a monochromatic emissivity  such that  = 0.20 from 0 to 4 m,  = 0.40 from 4 to 10 m, and  = 0.60 from 10 to ∞. Determine (a) the total emissivity and (b) the total emissive power. 26.41: The top surface of a circular metallic disk of 1-cm diameter is maintained at 900 K while the bottom surface is perfectly insulated. The disk is placed in an environment at 100 K. The monochromatic emissivity of the disk is 0.35 between  = 0 m and  = 2 m and 0.20 between  = 2 m and  = ∞. Determine the radiant energy (watts) emitted by the disk. 26.42: The monochromatic emissivity of a diffuse surface at 500 K is 0.65 between 0 and  and 0.52 between  and ∞; the total emissivity is 0.61. Determine the value of . 26.43: The monochromatic emissivity of a surface varies with wavelength as follows:  = 0.20 between  = 0 and 1 m,  = 0.50 between  = 1 and 6 m,  = 0.70 between  = 6 and 15 m, and  = 0.82 between  = 15 m and ∞. Determine the total emissive power of the surface at a temperature of 1200 K. 26.44: Assuming the sun to be a blackbody at 5800 K, determine the absorptivity of the surface in Problem 26.43 to solar radiation.

890

Introduction to Thermal and Fluid Engineering

26.45: The surface in Problem 26.42 receives an irradiation of 10,000 W/m2 originating from a blackbody at 2500 K. Determine (a) the radiant heat flux in W/m2 absorbed and (b) the reflected heat flux if the surface is opaque. The Shape Factor 26.46: In the arrangement of surfaces shown in Figure P26.46, surface 1 is a disk of 1-m diameter, surface 2 is a hemisphere of 2-m diameter, and surface 3, an imaginary surface, is shown dashed. Determine the shape factors, F12 , F21 , F23 , F32 , and F22 . 2 d2 = 2 m 1 d1 = 1 m 3

3 FIGURE P26.46

26.47: For the right circular cylinder of diameter d and length L , indicated in Figure P26.47, the end surfaces are designated as surfaces 1 and 2 and the lateral surface is designated as surface 3. If d = L, determine F12 and F13 . L

2

d

1 3 FIGURE P26.47

26.48: For a long duct with the cross section indicated in Figure P26.48, show that 1 F21 =  

− 2 S2 S1

S1

r θ

FIGURE P26.48

26.49: For inclined parallel plates of equal widths a and a common edge as indicated in Figure P26.49, prove that

F12 = 1 − sin 2

Radiation Heat Transfer

891

a S1 θ S2 a FIGURE P26.49

26.50: For the two-dimensional three-sided enclosure shown in Figure P26.50, show that the shape factor, F12 , is given by F12 =

a1 + a2 − a3 2a 1

a3

a2

S2

S1

S3 a1

FIGURE P26.50

26.51: For the two-dimensional perpendicular arrangement shown in Figure P26.51, establish the relation   2 1/2 a2 a2 − 1+ 1+ a1 a1 F12 = 2

a2

S2 S1 a1 FIGURE P26.51

26.52: Consider the truncated cone of height h and top and bottom radii of r1 and r2 , respectively, as shown in Figure P26.52. For the surface designations shown, determine the shape factors, F12 , F32 , and F22 . r1

3 2 1 r2 FIGURE P26.52

h

892

Introduction to Thermal and Fluid Engineering

26.53: In Figure P26.53, determine the shape factors F14 , F41 , F32 , and F23 .

2m 3m

2 1 3m

4 3 3m FIGURE P26.53

26.54: Determine the shape factors F12 and F21 for the configuration shown in Figure P26.54. 1m

2

1.2 m

1

1.5 m FIGURE P26.54

26.55: Consider the concentric cylinder arrangement shown in Figure P26.55 and derive the following expressions for the shape factors. 1 (1 − F12 − F11 ) 2   1 S1 = 1 − F12 2 S2

F12 = F23

F33 = 1 −

S1 + S2 S1 + (2F12 + F11 ) 2S3 2S2

Radiation Heat Transfer

893 S2 S3 S1

FIGURE P26.55

26.56: For the two-dimensional configuration shown in Figure P26.56, determine the shape factors F12 , F13 , and F14 .

0.8 m

1.5 m

2

1

3

4

0.4 m 0.1 m 0.2 m FIGURE P26.56

26.57: Determine the shape factors F12 and F21 for the two rectangles that lie in the mutually perpendicular planes indicated in Figure P26.57.

894

Introduction to Thermal and Fluid Engineering 0.40 m

1.20 m

1

2

0.30 m

0.50 m FIGURE P26.57

26.58: As indicated in Figure P26.58, the radiation from a cylindrical cavity of diameter 0.4 m and depth 0.8 m is measured by a circular sensor of 0.4 m placed parallel to the cavity opening at a distance of 0.3 m. Determine the fractions of the radiation received by the sensor (surface 3) from (a) the base of the cavity (surface 1) and (b) the lateral surfaces of the cavity (surface 2).

0.40 m

0.30 m

S3

0.80 m S2

S1 FIGURE P26.58

Heat Flow by Radiation between Two Bodies 26.59: Two black, infinitely long parallel surfaces at 1200 and 600 K face each other. Determine the net radiant heat exchange per unit surface area between the two surfaces. 26.60: Assume the surfaces in Figure P26.60 to be black, and determine the net radiant heat ˙ 12 , between the two aligned, parallel, rectangular surfaces shown. exchange, Q

Radiation Heat Transfer

895 0.20 m

T1 = 1500 K 0.50 m 1

0.30 m

2 T2 = 700 K FIGURE P26.60

26.61: Refer again to Problem 26.56. Assume all surfaces to be black, and take as the surface temperatures T1 = 1600 K and T2 = T3 = T4 = 500 K. Determine the net radiation ˙ 12 , Q ˙ 13 , and Q ˙ 14 . heat transfers, Q ˙ 12 when T1 = 500 K and T2 = 26.62: For the configuration of Figure P26.54, determine Q 1000 K if both surfaces are black. 26.63: Consider the concentric cylinder arrangement of Figure P26.55. Assuming surfaces ˙ 12 . Pertinent 1 and 2 to be black and at 1600 K and 600 K, respectively, determine Q dimensions are d1 = 1 m, d2 = 2 m, and L = 5 m. 26.64: Two gray, infinitely long, parallel surfaces at 1600 K and 400 K face each other. The emissivity of the hot surface is 0.90, while that of the cold surface is 0.50. Per unit of area, determine the net radiant energy heat exchange between the surfaces. 26.65: A thin radiation shield of emissivity 0.40 on both sides is inserted between the two gray surfaces described in Problem 26.64. Determine (a) the equilibrium temperature of the shield and (b) the percentage reduction in radiation heat exchange between the surfaces as a result of the shielding. 26.66: Two large gray parallel planes with T1 = 2000 K and 1 = 0.82 and T2 = 800 K and 2 = 0.75 face each other. The radiant heat exchange between the plates is to be reduced by 60% and the emissivity of the proposed radiation shield is the same on both sides. Determine (a) the emissivity of the shield and (b) the equilibrium temperature of the shield. 26.67: A thin radiation shield is placed between two parallel gray surfaces with T1 = 1400 K and 1 = 0.90 and T2 = 600 K and 2 = 0.70. The emissivities of the shield are 0.56 on the side facing the hot surface and 0.42 on the side facing the cooler surface. Determine the heat transfer per unit area (a) with the shield and (b) without the shield, and (c) determine the equilibrium temperature of the shield. 26.68: Consider two parallel gray surfaces of infinite extent with temperatures and emissivities of T1 = 1500 K and 1 = 0.80 and T2 = 600 K and 2 = 0.70. The two plates face each other and it is proposed that the heat transfer between the surfaces be reduced by a factor of 10 using radiation shields with emissivities of 0.40 on both sides. Determine the number of shields needed. 26.69: In a concentric cylinder arrangement, the outer surface of the inner cylinder (surface 1) has a diameter of 40 cm and the inner surface of the outer cylinder (surface 2) has a diameter of 80 cm. Temperatures and emissivities are T1 = 700 K and 1 = 0.75 and

896

Introduction to Thermal and Fluid Engineering T2 = 300 K and 2 = 0.65. Assuming that the cylinders are infinitely long, determine the rate of heat transfer between the hot and the cold surfaces.

26.70: For the arrangement in Problem 26.69, a thin cylindrical shield of 60-cm diameter and an emissivity of 0.25 is placed between the cylinders. Determine (a) the new rate of heat transfer between the two surfaces and (b) the equilibrium temperature of the shield. 26.71: For the concentric sphere arrangement indicated in Figure P26.71, show that the rate ˙ 12 , is given by of heat transfer, Q ˙ 12 = Q

4r12  1 (T14 − T24 )    2 1 1 r1 + −1

1

2 r2

(T1 > T2 )

T2, ε2 r2 r1

T1, ε1

FIGURE P26.71

Radiosity, Irradiation, and Radiation Networks 26.72: Two gray parallel plates of infinite extent with T1 = 1000 K and 1 = 0.75 and T2 = 400 K and 2 = 0.85 radiate to each other. Determine (a) the radiosities, J 1 and J 2 , (b) the irradiations, G 1 and G 2 , and (c) the rate of heat transfer per unit area. 26.73: A circular disk of radius 25 cm sits centrally at the base of a hemisphere of radius 50 cm as shown in Figure P26.73. Determine (a) the radiosities, J 1 and J 2 , (b) the ˙ 1 , and Q ˙ 2 , and (d) the heat irradiations G 1 and G 2 , (c) the rates of heat transfer, Q transfer to the environment at 280 K. 1 r1 = 50 cm T1 = 900 K ε1 = 0.70 2

r1 r2 r2 = 25 cm T2 = 400 K ε2 = 0.50

FIGURE P26.73

26.74: For the concentric arrangement in Problem 26.69 (see Figure P26.71) determine (a) the radiosities J 1 and J 2 , (b) the irradiations G 1 and G 2 , and (c) the net heat transfer ˙ 12 . Q 26.75: Consider the concentric sphere arrangement of Figure P26.71. With T1 = 800 K, r1 = 0.30 m, and 1 = 0.72 and T2 = 300 K, r2 = 0.60 m, and 2 = 0.62, determine (a) ˙ 12 . the radiosities J 1 and J 2 , (b) the irradiations G 1 and G 2 , and (c) Q

Radiation Heat Transfer

897

Radiation Networks 26.76: Two identical rectangular plates, each 30-cm wide and 50-cm high, face each other as shown in Figure P26.76. The plates sit in an environment at 0◦ C. Using the thermal network method, determine the net radiant heat exchange between the surfaces and the radiant heat loss from both plates to the environment areas. 20 cm 30 cm

T2 = 300 K ε2 = 0.80

T1 = 500 K ε1 = 0.70 50 cm

FIGURE P26.76

26.77: Consider the three-surface configuration shown in Figure P26.77. Draw the thermal network representing the configuration. The depth of all surfaces may be considered infinite and the surroundings may be considered to be at 0 K. T2 = 600 K ε2 = 0.72

T3 = 400 K ε3 = 0.65

0.10 m

3

2 1 0.40 m T1 = 800 K ε1 = 0.85 FIGURE P26.77

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Appendix A Tables and Charts

TABLE A.1

Values of the Gas Constant and Specific Heats for Several Gases

Gas

Molecular Weight, kg/kg mol

R at 300 K, kJ/kg-K

cp at 300 K, kJ/kg-K

cv at 300 K, kJ/kg-K

ka

28.97 39.94 44.01 28.01 30.07 4.003 2.016 16.04 32.00 28.02 44.09

0.287 0.208 0.189 0.297 0.276 2.078 4.127 0.519 0.260 0.297 0.188

1.005 0.515 0.846 1.041 1.767 5.234 14.32 2.227 0.917 1.038 1.692

0.718 0.310 0.653 0.745 1.495 3.146 10.16 1.735 0.653 0.741 1.507

1.400 1.661 1.296 1.400 1.182 1.664 1.409 1.284 1.404 1.401 1.123

Air Argon, A Carbon dioxide, CO2 Carbon monoxide, CO Ethane, C2 H6 Helium, He Hydrogen, H2 Methane, CH4 Oxygen, O2 Nitrogen, N2 Propane, C3 H8 a

Dimensionless parameter = c p /c v . Source: L. C. Nelson and E. F. Obert, Generalized Compressibility Charts. Chem. Eng., 61, 203 (1954).

TABLE A.2

Specific Heats and k = c p /c v for Six Gases as a Function of Absolute Temperature Air

N2

T, K

cp , kJ/kg-K

cv , kJ/kg-K

k

cp , kJ/kg-K

cv , kJ/kg-K

k

250 300 350 400 450 500 550 600 650 700 750 800 900 1000

1.003 1.005 1.008 1.013 1.020 1.029 1.040 1.051 1.063 1.075 1.087 1.099 1.121 1.142

0.726 0.718 0.721 0.726 0.733 0.742 0.753 0.764 0.776 0.788 0.800 0.812 0.834 0.855

1.401 1.400 1.398 1.395 1.391 1.387 1.381 1.376 1.370 1.364 1.359 1.354 1.344 1.336

1.039 1.039 1.041 1.044 1.049 1.056 1.065 1.075 1.086 1.098 1.110 1.121 1.145 1.167

0.742 0.743 0.744 0.747 0.752 0.759 0.768 0.778 0.789 0.801 0.813 0.825 0.849 0.870

1.400 1.400 1.399 1.397 1.385 1.391 1.387 1.382 1.376 1.371 1.365 1.360 1.349 1.341 (continued)

899

900

Appendix A

TABLE A.2 (Continued )

Specific Heats and k = c p /c v for Six Gases as a Function of Absolute Temperature O2

H2

T, K

cp , kJ/kg-K

cv , kJ/kg-K

k

cp , kJ/kg-K

cv , kJ/kg-K

k

250 300 350 400 450 500 550 600 650 700 750 800 900 1000

0.913 0.915 0.928 0.941 0.956 0.972 0.988 1.003 1.017 1.031 1.043 1.054 1.074 1.090

0.653 0.658 0.668 0.681 0.696 0.712 0.728 0.743 0.758 0.771 0.783 0.794 0.814 0.830

1.398 1.395 1.389 1.382 1.373 1.365 1.358 1.350 1.343 1.337 1.332 1.327 1.319 1.313

14.051 14.307 14.427 14.476 14.501 14.513 14.530 14.546 14.571 14.604 14.645 14.695 14.822 14.983

9.927 10.183 10,302 10.352 10.377 10.389 10.405 10.422 10.457 10.480 10.521 10.570 10.698 10.859

1.416 1.405 1.400 1.398 1.398 1.397 1.396 1.396 1.395 1.394 1.392 1.390 1.385 1.380

CO

CO2

T, K

cp , kJ/kg-K

Cv , kJ/kg-K

k

cp , kJ/kg-K

cv , kJ/kg-K

k

250 300 350 400 450 500 550 600 650 700 750 800 900 1000

1.039 1.040 1.043 1.047 1.054 1.063 1.075 1.087 1.100 1.113 1.126 1.139 1.163 1.185

0.743 0.744 0.746 0.751 0.757 0.767 0.778 0.790 0.803 0.816 0.829 0.842 0.866 0.868

1.400 1.399 1.398 1.395 1.392 1.387 1.382 1.376 1.370 1.364 1.358 1.353 1.343 1.335

0.791 0.846 0.895 0.939 0.978 1.014 1.046 1.075 1.102 1.126 1.148 1.169 1.204 1.234

0.602 0.657 0.706 0.750 0.790 0.825 0.857 0.886 0.913 0.937 0.969 0.980 1.015 1.045

1.314 1.288 1.268 1.252 1.239 1.229 1.220 1.213 1.207 1.202 1.197 1.193 1.186 1.181

Source: U.S. National Bureau of Standards, Tables of Thermal Properties of Gases, Circular 564, 1955.

0.61 0.87 1.23 1.71 2.34 3.17 4.24 5.62 7.37 9.58 12.33 15.74 19.92 25.00 31.15 38.54 47.35 57.80 70.10 84.52 101.32 120.80 143.27 169.07 198.55 232.11 270.15 313.09 361.39 415.53 475.99

vf

vfg

vg

0.001000 0.001000 0.001000 0.001001 0.001002 0.001003 0.001004 0.001006 0.001008 0.001010 0.001012 0.001014 0.001017 0.001020 0.001023 0.001026 0.001029 0.001032 0.001036 0.001039 0.001043 0.001047 0.001051 0.001056 0.001060 0.001065 0.001070 0.001075 0.001080 0.001085 0.001091

206.13 147.20 106.36 78.036 57.801 43.446 32.907 25.250 19.536 15.262 12.046 9.5771 7.6776 6.1996 5.0452 4.1328 3.4074 2.8276 2.3604 1.9806 1.6689 1.4142 1.2063 1.0350 0.89100 0.76938 0.66702 0.58074 0.50739 0.44462 0.39100

206.13 147.20 106.36 78.037 57.802 43.447 32.908 25.251 19.537 15.263 12.047 9.5781 7.6786 6.2006 5.0462 4.1338 3.4085 2.8286 2.3614 1.9817 1.6699 1.4152 1.2074 1.0361 0.8921 0.7704 0.6681 0.5818 0.5085 0.4457 0.3921

uf 0.00 21.04 42.02 62.95 83.86 104.75 125.63 146.50 167.37 188.24 209.12 230.01 250.91 271.83 292.76 313.70 334.67 355.65 376.66 397.69 418.75 439.83 460.95 482.10 503.28 524.51 545.78 567.09 588.46 609.88 631.35

ufg

ug

2,373.9 2,361.1 2,347.8 2,333.7 2,319.9 2,305.5 2,291.6 2,277.3 2,263.2 2,249.1 2,234.7 2,220.4 2,206.0 2,191.6 2,177.0 2,162.3 2,147.6 2,132.8 2,117.8 2,102.8 2,087.9 2,072.8 2,057.2 2,041.3 2,025.4 2,009.6 1,993.6 1,977.3 1,960.9 1,944.3 1,927.5

2,373.9 2,382.1 2,389.8 2,396.7 2,403.7 2,410.3 2,417.2 2,423.8 2,430.6 2,437.3 2,443.8 2,450.4 2,456.9 2,463.4 2,469.7 2,476.0 2,482.3 2,488.4 2,494.5 2,500.5 2,506.6 2,512.6 2,518.2 2,523.4 2,528.7 2,534.1 2,539.4 2,544.4 2,549.3 2,554.2 2,558.8

hf 0.0 21.0 42.0 63.0 83.9 104.8 125.6 146.5 167.4 188.3 209.1 230.0 250.9 271.9 292.8 313.7 334.7 355.7 376.7 397.8 418.9 440.0 461.1 482.3 503.5 524.8 546.1 567.4 588.8 610.3 631.9

Entropy, kJ/kg-K

hfg

hg

sf

sfg

sg

2,500.0 2,489.6 2,478.4 2,466.8 2,455.0 2,443.1 2,431.2 2,419.2 2,407.3 2,395.3 2,383.2 2,371.1 2,358.9 2,346.6 2,334.2 2,321.6 2,309.8 2,296.2 2,283.3 2,270.2 2,257.0 2,243.6 2,230.0 2,216.3 2,202.3 2,188.2 2,173.8 2,159.2 2,144.3 2,129.1 2,113.6

2,500.0 2,510.6 2,520.4 2,529.7 2,538.9 2,547.9 2,556.8 2,565.7 2,574.6 2,583.5 2,592.3 2,601.1 2,609.8 2,618.4 2,626.9 2,635.4 2,643.7 2,651.9 2,660.0 2,668.0 2,675.8 2,683.6 2,691.1 2,698.6 2,705.8 2,712.9 2,719.9 2,726.6 2,733.1 2,739.4 2,745.5

−0.0012 0.0757 0.1509 0.2244 0.2965 0.3672 0.4367 0.5049 0.5720 0.6381 0.7031 0.7672 0.8303 0.8926 0.9540 1.0146 1.0744 1.1335 1.1917 1.2493 1.3062 1.3624 1.4179 1.4728 1.5271 1.5807 1.6338 1.6864 1.7384 1.7899 1.8409

9.1590 8.9510 8.7511 8.5582 8.3718 8,1919 8.0180 7.8496 7.6864 7.5281 7.3745 7.2253 7.0804 6.9394 6.8023 6.6687 6.5387 6.4118 6.2881 6.1673 6.0492 5.9338 5.8209 5.7105 5.6023 5.4962 5.3922 5.2902 5.1900 5.0916 4.9948

9.1578 9.0267 8.9020 8.7827 8.6684 8.5591 8.4546 8.3545 8.2584 8.1662 8.0776 7.9925 7.9107 7.8320 7.7563 7.6834 7.6131 7.5453 7.4798 7.4166 7.3554 7.2962 7.2388 7.1833 7.1293 7.0770 7.0261 6.9766 6.9284 6.8815 6.8358

901

(continued )

88084˙Book

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 150

Psat , kPa

Enthalpy, kJ/kg

88084

Tsat C

Internal Energy, kJ/kg

13:51

Specific Volume, m3 /kg ,◦

P1: Gopal Joshi

March 22, 2011

Thermodynamic Properties of Steam: Temperature Table

Appendix A

TABLE A.3

88084

Specific Volume, m3 /kg Tsat C

vf

543.30 618.00 700.68 791.86 892.20 1,002.3 1,122.9 1,254.5 1,398.0 1,553.9 1,723.1 1,906.3 2,104.3 2,317.8 2,547.8 2,795.0 3,060.3 3,344.7 3,649.0 3,974.2 4,321.3 4,691.2 5,085.0 5,503.8 5,948.6

0.001096 0.001102 0.001108 0.001114 0.001121 0.001127 0.001134 0.001141 0.001148 0.001156 0.001164 0.001172 0.001180 0.001189 0.001198 0.001208 0.001218 0.001228 0.001239 0.001250 0.001262 0.001275 0.001288 0.001302 0.001317

vfg 0.34514 0.30566 0.27131 0.24141 0.21538 0.19266 0.17272 0.15513 0.13964 0.12597 0.11386 0.10307 0.09345 0.08486 0.07716 0.07022 0.06400 0.05851 0.05353 0.04893 0.04471 0.04086 0.03738 0.03424 0.03139

Enthalpy, kJ/kg

Entropy, kJ/kg-K

vg

uf

ufg

ug

hf

hfg

hg

0.3462 0.3068 0.2724 0.2425 0.2165 0.1938 0.1739 0.1563 0.1408 0.1271 0.1150 0.1042 0.0946 0.0860 0.0784 0.0714 0.0652 0.0597 0.0548 0.0502 0.0460 0.0421 0.0387 0.0355 0.0327

652.89 674.50 696.18 717.93 739.77 761.69 783.70 805.80 828.01 850.32 872.74 895.28 917.94 940.73 963.66 986.73 1,010.0 1,033.6 1,056.9 1,080.7 1,104.6 1,128.8 1,153.2 1,177.9 1,202.8

1,910.3 1,892.8 1,875.1 1,857.2 1,839.0 1,820.5 1,803.6 1,782.4 1,762.9 1,743.0 1,722.7 1,702.0 1,681.0 1,659.5 1,637.6 1,615.4 1,592,7 1,569.1 1,545.2 1,521.0 1,496.7 1,471.9 1,446.4 1,420.3 1,393.4

2,563.2 2,567.3 2,571.3 2,575.2 2,578.8 2,582.1 2,585.3 2,588.3 2,590.9 2,593.3 2,595.4 2,597.3 2,598.9 2,600.2 2,601.3 2,602.2 2,602.7 2,602.5 2,602.1 2,601.7 2,601.3 2,600.7 2,599.6 2,598.1 2,596.3

653.5 675.2 697.0 718.8 740.8 762.8 785.0 807.2 829.6 852.1 874.7 897.5 920.4 943.5 966.7 990.1 1,033.7 1,037.5 1,061.4 1,085.6 1,110.1 1,134.8 1,159.8 1,185.1 1,210.7

2,097.8 2,081.7 2,065.2 2,048.4 2,031.2 2,013.6 1,995.5 1,977.1 1,958.1 1,938.8 1,918.9 1,898.5 1,877.6 1,856.2 1,834.2 1,811.7 1,788.6 1,764.8 1,740.5 1,715.5 1,689.9 1,663.5 1,636.5 1,608.7 1,580.1

2,751.3 2,756.9 2,762.2 2,767.2 2,772.0 2,776.4 2,780.5 2,784.3 2,787.8 2,790.9 2,793.6 2,796.0 2,798.0 2,799.7 2,800.9 2,801.8 2,802.3 2,802.3 2,801.9 2,801.2 2,800.0 2,798.3 2,796.3 2,793.7 2,790.8

sf 1.8915 1.9416 1.9912 2.0405 2.0894 2.1380 2.1862 2.2341 2.2817 2.3290 2.3761 2.4230 2.4696 2.5161 2.5623 2.6084 2.6544 2.7002 2.7460 2.7917 2,8373 2.8829 2.9286 2.9743 3.0200

sfg

sg

4.8996 4.8059 4.7135 4.6224 4.5325 4.4437 4.3559 4.2691 4.1834 4.0986 4.0147 3.9314 3.8485 3.7661 3.6841 3.6025 3.5213 3.4404 3.3597 3.2792 3.1986 3.1180 3.0372 2.9560 2.8745

6.7911 6.7475 6.7048 6.6630 6.6220 6.5817 6.5421 6.5032 6.4651 6.4276 6.3908 6.3544 6.3181 6.2821 6.2464 6.2109 6.1757 6.1406 6.1057 6.0708 6.0359 6.0009 5.9657 5.9303 5.8945

Appendix A

155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275

Psat , kPa

Internal Energy, kJ/kg

88084˙Book

,◦

13:51

Thermodynamic Properties of Steam: Temperature Table

P1: Gopal Joshi

March 22, 2011

902

TABLE A.3 (Continued )

0.0301 0.0277 0.0255 0.0235 0.0217 0.0199 0.0183 0.0169 0.0155 0.0142 0.0130 0.0119 0.0108 0.0098 0.0088 0.0079 0.0069 0.0060 0.0050 0.00315

1,228.1 1,253.7 1,279.6 1,306.0 1,332.8 1,360.2 1,388.0 1,416.5 1,445.7 1,475.5 1,506.2 1,537.8 1,570.4 1,606.3 1,643.0 1,682.1 1,726.2 1,777.9 1,843.3 2,029.6

1,366.0 1,337.9 1,297.7 1,265.5 1,232.0 1,197.5 1,161.1 1,123.0 1,083.0 1,040.9 996.3 949.0 898.7 842.1 780.5 710.9 629.5 531.0 394.1 0.0

2,594.0 2,591.6 2,577.3 2,571.5 2,564.8 2,557.6 2,549.2 2,539.6 2,528.7 2,516.4 2,502.5 2,486.8 2,469.0 2,448.4 2,423.5 2,393.0 2,355.7 2,308.9 2,237.3 2,029.6

1,236.6 1,263.0 1,289.8 1,317.1 1,344.9 1,373.3 1,402.3 1,432.1 1,462.6 1,494.0 1,526.3 1,559.7 1,594.3 1,632.5 1,671.8 1,713.9 1,761.6 1,817.8 1,890.1 2,099.3

1,550.8 1,520.6 1,477.9 1,442.8 1,406.2 1,367.9 1,327.8 1,285.5 1,240.9 1,193.6 1,143.3 1,089.6 1,032.2 967.7 897.2 817.3 723.7 610.3 451.9 0.0

2,787.4 2,783.6 2,767.7 2,759.9 2,751.1 2,741.2 2,730.1 2,717.6 2,703.5 2,687.3 2,669.6 2,649.3 2,626.5 2,600.2 2,569.0 2,531.2 2,485.3 2,428.1 2,342.0 2,099.3

3.0660 3.1121 3.1585 3.2052 3.2523 3.3000 3.3483 3.3973 3.4473 3.4984 3.5507 3.6045 3.6601 3.7176 3.7775 3.8400 3.9056 3.9746 4.0476 4.4298

2.7924 2.7097 2.6262 2.5417 2.4560 2.3688 2.2797 2.1884 2.0947 1.9979 1.8973 1.7922 1.6820 1.5658 1.4416 1.3054 1.1531 0.9822 0.7555 0.0

88084˙Book

0.02878 0.02639 0.02418 0.02214 0.02025 0.01850 0.01688 0.01538 0.01398 0.01267 0.01143 0.01026 0.00914 0.00808 0.00706 0.00605 0.00504 0.00400 0.00274 0.00000

88084

0.001333 0.001349 0.001366 0.001385 0.001404 0.001425 0.001447 0.001470 0.001499 0.001528 0.001561 0.001598 0.001639 0.001686 0.001741 0.001808 0.001896 0.002016 0.002225 0.00315

13:51

6,420.5 6,920.8 7,450.6 8,011.1 8,603.7 9,214.4 9,869.4 10,561. 11,289. 12,056. 12,862. 13,712. 14,605. 15,545. 16,535. 17,577. 18,675. 19,833. 21,054. 22,090.

5.8584 5.8218 5.7847 5.7469 5.7083 5.6687 5.6279 5.5858 5.5420 5.4962 5.4480 5.3967 5.3420 5.2834 5.2191 5.1454 5.0587 4.9569 4.8030 4.4298

Source: Properties obtained from software, STEAMCALC. John Wiley & Sons, New York, 1983.

P1: Gopal Joshi

March 22, 2011

Appendix A

280 285 290 295 300 305 310 315 320 325 330 335 340 345 350 355 360 365 370 374.4

903

Specific Volume, m3 /kg vfg

vg

0.001000 0.001001 0.001001 0.001002 0.001003 0.001003 0.001004 0.001005 0.001005 0.001006 0.001006 0.001007 0.001007 0.001008 0.001008 0.001009 0.001009 0.001010 0.001010 0.001014 0.001017 0.001020 0.001022 0.001024 0.001026 0.001028 0.001030 0.001033 0.001036 0.001038 0.001041 0.001043 0.001043

129.08 88.067 67.073 54.290 45.751 39.483 34.779 31.128 28.194 25.773 23.742 22.013 20.522 19.225 18.086 17.080 16.185 15.383 14.660 10.020 7.6483 6.2015 5.2277 4.5249 3.9918 3.5744 3.2389 2.7305 2.3638 2.0859 1.8667 1.6898 1.66895

129.08 88.068 67.074 54.291 45.752 39.484 34.780 31.129 28.195 25.774 23.743 22.014 20.523 19.226 18.087 17.081 16.186 15.384 14.661 10.021 7.6493 6.2025 5.2287 4.5259 3.9929 3.5755 3.2398 2.7316 2.3648 2.0869 1.8678 1.6908 1.6700

uf 29.40 54.68 73.41 88.41 100.96 111.81 121.37 129.95 137.73 144.86 151.45 157.58 163.31 168.70 173.79 178.61 183.19 187.56 191.74 225.83 251.28 271.80 289.09 304.11 317.42 329.40 340.31 359.66 376.49 391.43 404.90 417.20 418.74

ufg

ug

2,356.1 2,339.3 2,326.8 2,316.7 2,308.0 2,300.9 2,294.5 2,288.6 2,283.3 2,278.4 2,274.0 2,269.8 2,266.0 2,262.3 2,258.9 2,255.6 2,252.5 2,249.5 2,246.7 2,223.2 2,205.7 2,191.6 2,379.5 2,169.0 2,159.7 2,151.3 2,143.6 2,129.9 2,117.9 2,107.2 2,097.7 2,089.0 2,087.9

2,385.5 2,394.0 2,400.2 2,405.1 2,409.0 2,412.7 2,415.9 2,418.6 2,421.0 2,423.3 2,425.4 2,427.4 2,429.3 2,431.0 2,432.7 2,434.2 2,435.7 2,437.1 2,438.4 2,449.0 2,457.0 2,463.3 2,468.6 2,473.1 2,477.1 2,480.7 2,483.9 2,489.6 2,494.4 2,498.7 2,502.6 2,506.2 2,506.6

hf 29.4 54.7 73.4 88.4 101.0 111.8 121.4 130.0 137.7 144.9 151.5 157.6 163.3 168.7 173.8 178.6 183.2 187.6 191.7 225.8 251.3 271.8 289.1 304.1 317.5 329.4 340.4 359.7 376.6 391.5 405.0 417.3 418.8

Entropy, kJ/kg-K

hfg

hg

sf

sfg

sg

2,485.2 2,471.4 2,460.9 2,452.4 2,445.3 2,439.1 2,433.6 2,428.7 2,424.3 2,420.2 2,416.4 2,412.9 2,409.6 2,406.5 2,403.6 2,400.8 2,398.2 2,395.7 2,393.3 2,373.5 2,358.7 2,346.6 2,336.3 2,327.4 2,319.4 2,312.2 2,305.5 2,293.7 2,283.4 2,274.1 2,265.7 2,258.0 2,257.0

2,514.6 2,526.3 2,534.3 2,540.8 2,546.2 2,550.9 2,555.0 2,558.7 2,562.0 2,565.0 2,567.9 2,570.5 2,572.9 2,575.2 2,577.4 2,579.4 2,581.4 2,583.2 2,585.0 2,599.4 2,610.0 2,618.4 2,625.5 2,631.5 2,636.8 2,641.6 2,645.9 2,653.5 2,659.9 2,665.6 2,670.7 2,675.3 2,675.8

0.1058 0.1956 0.2607 0.3120 0.3545 0.3908 0.4226 0.4509 0.4764 0.4996 0.5209 0.5407 0.5590 0.5763 0.5924 0.6077 0.6222 0.6359 0.6490 0.7544 0.8314 0.8925 0.9433 0.9869 1.0253 1.0595 1.0904 1.1446 1.1913 1.2323 1.2689 1.3020 1.3062

8.8704 8.6337 8.4642 8.3322 8.2240 8.1324 8.0529 7.9827 7.9197 7.8626 7.8104 7.7623 7.7177 7.6762 7.6372 7.6006 7.5660 7.5332 7.5021 7.2548 7.0779 6.9396 6.8260 6.7295 6.6455 6.5710 6.5042 6.3880 6.2891 6.2029 6.1265 6.0578 6.0493

8.9763 8.8292 8.7249 8.6442 8.5786 8.5233 8.4756 8.4336 8.3961 8.3622 8.3313 8.3030 8.2768 8.2524 8.2296 8.2083 8.1881 8.1691 8.1511 8.0092 7.9093 7.8321 7.7693 7.7164 7.6707 7.6305 7.5946 7.5326 7.4804 7.4352 7.3954 7.3598 7.3554

88084˙Book

7.0 13.0 17.5 21.1 24.1 26.7 29.0 31.0 32.9 34.6 36.2 37.7 39.0 40.3 41.5 42.7 43.8 44.8 45.8 54.0 60.1 65.0 69.1 72.7 75.9 78.7 81.3 86.0 90.0 93.5 96.7 99.6 100.0

vf

Enthalpy, kJ/kg

Appendix A

1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 60.00 70.00 80.00 90.00 100.00 101.32

Tsat

Internal Energy, kJ/kg

88084

Psat , kPa

,◦ C

13:51

Thermodynamic Properties of Steam: Pressure Table

P1: Gopal Joshi

March 22, 2011

904

TABLE A.4

444.01 466.74 486.58 504.25 520.22 534.82 548.30 560.83 572.57 583.60 594.03 603.93 613.36 622.35 630.97 639.24 654.86 669.42 683.06 695.93 708.11 719.68 730.71 741.27 751.39 761.11 779.52 796.71 812.87 828.13 842.61 856.39 869.56 882.18 894.30 905.97

2,069.7 2,052.9 2,037.9 2,024.7 2,012.8 2,001.9 1,991.7 1,982.1 1,973.1 1,964.6 1,956.6 1,948.9 1,941.6 1,934.5 1,927.7 1,921.2 1,908.7 1,896.9 1,885.8 1,875.3 1,865.3 1,855.7 1,846.5 1,837.7 1,829.1 1,820.8 1,805.1 1,790.2 1,776.1 1,762.7 1,749.8 1,737.3 1,725.4 1,713.9 1,702.7 1,691.9

2,513.7 2,519.6 2,524.5 2,529.0 2,533.0 2,536.7 2,540.0 2,542.9 2,545.7 2,548.2 2,550.6 2,552.8 2,554.9 2,556.9 2,558.7 2,560.4 2,563.5 2,566.3 2,568.8 2,571.2 2,573.4 2,575.4 2,577.2 2,578.9 2,580.5 2,581.9 2,584.6 2,586.9 2,589.0 2,590.8 2,592.4 2,593.7 2,595.0 2,596.1 2,597.0 2,597.9

444.1 466.9 486.8 504.5 520.5 535.1 548.6 561.2 572.9 584.0 594.4 604.4 613.8 622.8 631.5 639.8 655.5 670.1 683.8 696.7 708.9 720.6 731.7 742.3 752.5 762.2 780.8 798.1 814.4 829.7 844.3 858.2 871.5 884.3 896.5 908.3

2,240.9 2,226.3 2,213.4 2,201.7 2,191.1 2,181.2 2,172.1 2,163.5 2,155.4 2,147.7 2,140.3 2,133.3 2,126.6 2,120.1 2,113.9 2,107.9 2,096.4 2,085.5 2,075.2 2,065.4 2,056.0 2,047.0 2,038.4 2,030,0 2,021.9 2,014.0 1,999.0 1,984.7 1,971.1 1,958.0 1,945.5 1,933.4 1,921.7 1,910.4 1,899.4 1,888.7

2,685.1 2,693.2 2,700.1 2,706.2 2,711.5 2,716.3 2,720.7 2,724.6 2,728.3 2,731.6 2,734.8 2,737.7 2,740.4 2,743.0 2,745.4 2,747.6 2,751.8 2,755.6 2,759.0 2,762.1 2,765.0 2,767.6 2,770.0 2.772.3 2,774.3 2,776.3 2,779.8 2,782.8 2,785.4 2,787.8 2,789.8 2,791.7 2,793.3 2,794.7 2,795.9 2,797.0

1.3734 1.4330 1.4844 1.5295 1.5700 1.6066 1.6401 1.6710 1.6998 1.7266 1.7519 1.7757 1.7982 1.8196 1.8400 1.8595 1.8960 1.9298 1.9613 1.9907 2.0183 2.0445 2.0692 2.0928 2.1152 2.1367 2.1771 2.2145 2.2493 2.2820 2.3127 2.3418 2.3695 2.3958 2.4210 2.4450

5.9113 5.7904 5.6873 5.5974 5.5175 5.4455 5.3800 5.3199 5.2643 5.2126 5.1642 5.1187 5.0758 5.0352 4.9966 4.9598 4.8910 4.8278 4.7692 4.7146 4.6634 4.6152 4.5696 4.5264 4.4853 4.4460 4.3725 4.3047 4.2417 4.1829 4.1277 4.0757 4.0265 3.9797 3.9350 3.8922

7.2847 7.2234 7.1717 7.1269 7.0874 7.0521 7.0201 6.9910 6.9641 6.9392 6.9161 6.8944 6.8740 6.8548 6.8366 6.8193 6.7871 6.7576 6.7305 6.7053 6.6817 6.6597 6.6388 6.6192 6.6005 6.5827 6.5495 6.5191 6.4910 6.4649 6.4405 6.4176 6.3960 6.3755 6.3559 6.3372

905

(continued )

88084˙Book

1.3707 1.1572 1.0035 0.8860 0.7934 0.7186 0.6571 0.6056 0.5619 0.5241 0.4911 0.4621 0.4364 0.4135 0.3930 0.3745 0.3424 0.3155 0.2926 0.2728 0.2555 0.2403 0.2268 0.2148 0.2041 0.1943 0.1774 0.1632 0.1511 0.1407 0.1317 0.1237 0.1167 0.1104 0.1047 0.0996

88084

1.36965 1.15612 1.00248 0.88498 0.79229 0.71751 0.65602 0.60457 0.56082 0.52305 0.49007 0.46105 0.43534 0.41242 0.39188 0.37336 0.34129 0.31443 0.29151 0.27168 0.25439 0.23919 0.22572 0.21372 0.20295 0.19322 0.17631 0.16209 0.14998 0.13956 0.13050 0.12254 0.11549 0.10918 0.10351 0.09839

13:51

0.001048 0.001053 0.001057 0.001060 0.001064 0.001067 0.001070 0.001073 0.001076 0.001079 0.001081 0.001084 0.001086 0.001088 0.001091 0.001093 0.001097 0.001101 0.001104 0.001108 0.001111 0.001115 0.001118 0.001121 0.001124 0.001127 0.001133 0.001138 0.001143 0.001148 0.001153 0.001158 0.001163 0.001167 0.001172 0.001176

P1: Gopal Joshi

March 22, 2011

106.0 111.4 116.1 120.2 124.0 127.4 130.6 133.5 136.3 138.9 141.3 143.6 145.8 147.9 149.9 151.8 155.5 158.8 162.0 164.9 167.7 170.4 172.9 175.3 177.7 179.9 184.1 187.9 191.6 195.0 198.3 201.4 204.3 207.1 209.8 212.4

Appendix A

125.00 150.00 175.00 200.00 225.00 250.00 275.00 300.00 325.00 350.00 375.00 400.00 425.00 450.00 475.00 500.00 550.00 600.00 650.00 700.00 750.00 800.00 850.00 900.00 950.00 1,000 1,100 1,200 1,300 1,400 1,500 1,600 1,700 1,800 1,900 2,000

Psat , kPa

0.001186 0.001196 0.001206 0.001215 0.001225 0.001234 0.001242 0.001251 0.001285 0.001319 0.001352 0.001385 0.001418 0.001452 0.001489 0.001527 0.001568 0.001612 0.001661 0.001715 0.001769 0.001839 0.001926 0.002037 0.002208 0.002623 0.00315

vfg 0.08751 0.07873 0.07147 0.06538 0.06028 0.05594 0.05208 0.04864 0.03811 0.03109 0.02603 0.02214 0.01907 0.01658 0.01450 0.01273 0.01119 0.00984 0.00862 0.00752 0.00660 0.00566 0.00475 0.00384 0.00281 0.00114 0.00000

vg 0.0887 0.0799 0.0727 0.0666 0.0615 0.0572 0.0533 0.0499 0.0394 0.0324 0.0274 0.0235 0.0205 0.0180 0.0160 0.0143 0.0128 0.0114 0.0103 0.0092 0.0084 0.0075 0.0067 0.0059 0.0050 0.0038 0.00315

Internal Energy, kJ/kg

Enthalpy, kJ/kg

Entropy, kJ/kg-K

uf

ufg

ug

hf

hfg

hg

sf

sfg

sg

933.41 958.75 982.37 1,004.5 1,025.4 1,045.3 1,064.2 1,082.2 1,147.9 1,205.6 1,257.8 1,305.9 1,351.0 1,393.7 1,434.6 1,474.0 1,512.4 1,549.8 1,586.6 1,626.1 1,660.2 1,698.6 1,740.1 1,785.8 1,839.7 1,944.6 2,029.6

1,666.2 1,642.1 1,619.4 1,597.9 1,577.0 1,556.8 1,537.6 1,519.3 1,451.9 1,390.4 1,333.5 1,265.8 1,209.3 1,153.8 1,098.6 1,043.3 987.6 931.1 873.4 810.1 750.3 680.8 603.6 515.1 401.7 174.0 0.0

2,599.6 2,600.9 2,601.8 2,602.4 2,602.5 2,602.1 2,601.8 2,601.5 2,599.8 2,596.0 2,591.3 2,571.7 2,560.3 2,547.5 2,533.1 2,517.3 2,499.9 2,480.9 2,460.0 2,436.2 2,410.5 2,379.4 2,343.6 2,300.9 2,241.4 2,118.6 2,029.6

936.1 961.7 985.7 1,008.2 1,029.4 1,049.6 1,068.8 1,087.2 1,154.3 1,213.5 1,267.2 1,317.0 1,363.7 1,408.2 1,451.0 1,492.4 1,532.8 1,572.4 1,611.5 1,653.6 1,690.3 1,731.7 1,776.7 1,826.6 1,886.0 2,002.3 2,099.3

1,863.1 1,839.0 1,816.0 1,794.0 1,772.9 1,752.6 1,732.9 1,713.9 1,642.4 1,576.9 1,515.7 1,442.9 1,380.9 1,319.5 1,258.0 1,196.0 1,133.1 1,068.8 1,002.8 930.4 862.5 782.6 693.9 591.9 460.8 199.0 0.0

2,799.2 2,800.7 2,801,7 2,802.2 2,802.3 2,802.2 2,801.7 2,801.1 2,796.7 2,790.5 2,782.9 2,759.9 2,744.7 2,727.7 2,709.0 2,688.4 2,665.8 2,641.2 2,614.3 2,584.0 2,552.8 2,514.3 2,470.5 2,418.5 2,346.8 2,201.3 2,099.3

2.5012 2.5525 2.5997 2.6437 2.6848 2.7234 2.7600 2.7947 2.9186 3.0251 3.1194 3.2050 3.2840 3.3580 3.4283 3.4958 3.5611 3.6249 3.6877 3.7498 3.8054 3.8652 3.9249 3.9846 4.0443 4.1042 4.4298

3.7925 3.7015 3.6178 3.5402 3.4676 3.3995 3.3350 3.2739 3.0548 2.8655 2.6965 2.5421 2.3981 2.2616 2.1305 2.0028 1.8771 1.7520 1.6265 1.4996 1.3819 1.2481 1.1063 0.9568 0.7681 0.4563 0.0

6.2936 6.2540 6.2176 6.1839 6.1524 6.1229 6.0950 6.0686 5.9734 5.8906 5.8159 5.7471 5.6821 5.6196 5.5588 5.4986 5.4382 5.3769 5.3142 5.2494 5.1872 5.1134 5.0312 4.9414 4.8124 4.5605 4.4298

Source: Properties obtained from software, STEAMCALC. John Wiley & Sons, New York, 1983.

88084˙Book

218.4 223.9 229.1 233.8 238.3 242.5 246.5 250.3 263.9 275.5 285.8 295.0 303.3 311.0 318.1 324.7 331.0 336.9 342.4 347.7 352.3 357.0 361.4 365.7 369.8 373.7 374.4

vf

Appendix A

2,250 2,500 2,750 3,000 3,250 3,500 3,750 4,000 5,000 6,000 7,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 22,090

Tsat

88084

Specific Volume, m3 /kg ,◦ C

13:51

Thermodynamic Properties of Steam: Pressure Table

P1: Gopal Joshi

March 22, 2011

906

TABLE A.4 (Continued )

Appendix A

907

TABLE A.5

Thermodynamic Properties of Steam: Superheated Vapor Table P, kPa

T,

◦C

10 (Tsat = 45.8◦ C) 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 50 (Tsat = 81.3◦ C) 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 100 (Tsat = 99.6◦ C) 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850

v, m3 /kg

17.196 19.513 21.826 24.136 26.446 28.754 31.063 33.371 35.679 37.987 40.295 42.603 44.911 47.219 49.526 51.834

3.4182 3.8894 4.3561 4.8206 5.2840 5.7468 6.2092 6.6715 7.1336 7.5956 8.0575 8.5193 8.9811 9.4428 9.9045 10.366

1.6956 1.9363 2.1724 2.4062 2.6388 2.8708 3.1025 3.3340 3.5654 3.7966 4.0277 4.2588 4.4898 4.7208 4.9518 5.1827

u, kJ/kg

2516.2 2588.2 2661.2 2735.5 2811.2 2888.3 2967.1 3047.4 3129.4 3213.1 3298.5 3385.7 3474.5 3565.2 3657.6 3751.8

2512.0 2586.0 2659.8 2734.5 2810.5 2887.8 2966.6 3047.1 3129.1 3212.9 3298.3 3385.5 3474.4 3565.0 3657.5 3751.7

2506.4 2583.1 2658.1 2733.3 2809.6 2887.1 2966.1 3046.6 3128.8 3212.5 3298.0 3385.2 3474.1 3564.8 3657.3 3751.5

h, kJ/kg

2688.1 2783.3 2879.5 2976.8 3075.6 3175.9 3277.7 3381.1 3486.2 3593.0 3701.5 3811.7 3923.7 4037.4 4152.9 4270.2

2682.9 2780.5 2877.6 2975.6 3074.7 3175.1 3277.1 3380.6 3485.8 3592.6 3701.2 3811.4 3923.4 4037.2 4152.7 4270.0

2676.0 2776.8 2875.3 2974.0 3073.5 3174.2 3276.4 3380.0 3485.3 3592.2 3700.8 3811.1 3923.1 4036.9 4152.4 4269.8

s, kJ/kg-K

8.4498 8.6893 8.9040 9.0996 9.2799 9.4476 9.6048 9.7530 9.8935 10.027 10.155 10.278 10.396 10.510 10.620 10.727

7.6959 7.9413 8.1583 8.3551 8.5360 8.7040 8.8614 9.0098 9.1504 9.2843 9.4123 9.5351 9.6532 9.7672 9.8775 9.9843

7.3610 7.6146 7.8347 8.0329 8.2146 8.3831 8.5407 8.6893 8.8300 8.9640 9.0921 9.2149 9.3331 9.4471 9.5574 9.6643

P, kPa

T,

v, m3 /kg

u, kJ/kg

h, kJ/kg

s, kJ/kg-K

101.32 (Tsat = 100.0◦ C) 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850

1.9109 2.1439 2.3747 2.6043 2.8334 3.0621 3.2905 3.5189 3.7471 3.9752 4.2033 4.4313 4.6593 4.8872 5.1152

2583.1 2658.0 2733.3 2809.6 2887.1 2966.1 3046.6 3128.7 3212.5 3298.0 3385.2 3474.1 3564.8 3657.3 3751.5

2776.7 2875.3 2973.9 3073.4 3174.2 3276.3 3380.0 3485.3 3592.2 3700.8 3811.1 3923.1 4036.9 4152.4 4269.8

7.6084 7.8286 8.0268 8.2085 8.3770 8.5347 8.6832 8.8240 8.9580 9.0860 9.2089 9.3271 9.4411 9.5513 9.6582

200 (Tsat = 120.2◦ C) 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850

0.9596 1.0804 1.1989 1.3162 1.4329 1.5492 1.6653 1.7813 1.8971 2.0129 2.1286 2.2442 2.3598 2.4754 2.5909

2577.2 2654.5 2730.9 2807.8 2885.8 2965.0 3045.7 3128.0 3211.9 3297.4 3384.7 3473.7 3564.4 3656.9 3751.1

2769.1 2870.6 2970.7 3071.1 3172.4 3274.9 3378.8 3484.3 3591.3 3700.0 3810.4 3922.5 4036.4 4152.0 4269.3

7.2804 7.5072 7.7084 7.8916 8.0610 8.2192 8.3682 8.5092 8.6434 8.7716 8.8945 9.0128 9.1268 9.2372 9.3441

300 (Tsat = 133.5◦ C) 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850

0.6338 0.7164 0.7965 0.8753 0.9535 1.0314 1.1091 1.1866 1.2639 1.3413 1.4185 1.4957 1.5728 1.6499 1.7270

2570.8 2650.8 2728.5 2806.1 2884.4 2963.9 3044.8 3127.2 3211.2 3296.9 3384.2 3473.2 3564.0 3656.5 3750.8

2760.9 2865.7 2967.4 3068.7 3170.5 3273.4 3377.6 3483.2 3590.4 3699.2 3809.7 3921.9 4035.8 4151.5 4268.9

7.0779 7.3122 7.5165 7.7014 7.8717 8.0305 8.1798 8.3211 8.4554 8.5838 8.7068 8.8252 8.9393 9.0497 9.1566

◦C

(continued )

908

Appendix A

TABLE A.5 (Continued )

Thermodynamic Properties of Steam: Superheated Vapor Table

UFM|484942|1435612122

P, kPa

T,

◦C

400 (Tsat =143.6◦ C) 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 600 (Tsat =158.8◦ C) 200 250 300 350 400 450 500 550 600 650 700 750 800 850 800 (Tsat =170.4◦ C) 200 250 300 350 400 450 500 550 600 650 700 750 800 850

v, m3 /kg

0.4707 0.5343 0.5952 0.6549 0.7139 0.7725 0.8309 0.8892 0.9474 1.0054 1.0635 1.1214 1.1793 1.2372 1.2951

0.3521 0.3939 0.4344 0.4742 0.5136 0.5528 0.5919 0.6308 0.6696 0.7084 0.7471 0.7858 0.8245 0.8631

0.2608 0.2932 0.3241 0.3544 0.3842 0.4137 0.4432 0.4725 0.5017 0.5309 0.5600 0.5891 0.6181 0.6471

u, kJ/kg

2563.9 2647.0 2726.0 2804.3 2883.1 2962.9 3044.0 3126.5 3210.6 3296.3 3383.7 3472.7 3563.5 3656.1 3750.4

2639.0 2720.8 2800.6 2880.3 2960.7 3042.2 3125.0 3209.3 3295.1 3382.6 3471.8 3562.7 3655.3 3749.7

2630.4 2715.5 2796.9 2877.5 2958.5 3040.4 3123.5 3208.0 3294.0 3381.6 3470.9 3561.9 3654.6 3749.0

h, kJ/kg

2752.2 2860.7 2964.1 3066.2 3168.6 3271.9 3376.3 3482.2 3589.5 3698.5 3809.0 3921.3 4035.3 4151.0 4268.4

2850.2 2957.2 3061.3 3164.8 3268.9 3373.8 3480.1 3587.7 3696.9 3807.7 3920.1 4034.2 4150.0 4267.6

2839.1 2950.1 3056,2 3161.0 3265.8 3371.4 3478.0 3586.0 3695.4 3806.3 3918.9 4033.1 4149.1 4266.7

s, kJ/kg-K

6.9287 7.1712 7.3789 7.5655 7.7367 7.8961 8.0458 8.1873 8.3219 8.4504 8.5735 8.6919 8.8062 8.9166 9.0236

6.9669 7.1819 7.3719 7.5451 7.7057 7.8562 7.9982 8.1332 8.2619 8.3853 8.5039 8.6182 8.7287 8.8358

6.8156 7.0388 7.2326 7.4079 7.5697 7.7209 7.8635 7.9988 8.1278 8.2514 8.3702 8.4846 8.5953 8.7024

P, kPa

T,

v, m3 /kg

u, kJ/kg

h, kJ/kg

s, kJ/kg-K

1000 (Tsat =179.9◦ C) 200 250 300 350 400 450 500 550 600 650 700 750 800 850

0.2059 0.2328 0.2580 0.2824 0.3065 0.3303 0.3540 0.3775 0.4010 0.4244 0.4477 0.4710 0.4943 0.5175

2621.4 2710.0 2793.1 2874.7 2956.3 3038.5 3121.9 3206.6 3292.8 3380.6 3470.0 3561.0 3653.8 3748.3

2827.3 2942.8 3051.1 3157.1 3262.7 3368.8 3475.9 3584.2 3693.8 3805.0 3917.7 4032.0 4148.1 4265.8

6.6930 6.9251 7.1229 7.3003 7.4633 7.6154 7.7585 7.8942 8.0235 8.1473 8.2662 8.3808 8.4916 8.5988

1500 (Tsat =198.3◦ C) 250 300 350 400 450 500 550 600 650 700 750 800 850

0.15199 0.16971 0.18654 0.20292 0.21906 0.23503 0.25089 0.26666 0.28237 0.29803 0.31364 0.32921 0.34475

2695.4 2783.3 2867.4 2950.6 3034.0 3118.1 3203.3 3289.9 3378.0 3467.6 3558.9 3651.8 3746.5

2923.4 3037.8 3147.2 3255.0 3362.5 3470.6 3579.7 3689.9 3801.5 3914.7 4029.4 4145.7 4263.6

6.7093 6.9183 7.1014 7.2677 7.4219 7.5664 7.7030 7.8331 7.9574 8.0767 8.1917 8.3027 8.4101

2000 (Tsat = 212.4◦ C) 250 300 350 400 450 500 550 600 650 700 750 800 850

0.11145 0.12550 0.13856 0.15113 0.16343 0.17556 0.18757 0.19950 0.21137 0.22318 0.23494 0.24667 0.25836

2679.5 2772.9 2860.0 2944.8 3029.3 3114.2 3200.0 3287.0 3375.4 3465.3 3556.8 3649.9 3744.7

2902.4 3023.9 3137.1 3247.1 3356.1 3465.3 3575.1 3686.0 3798.1 3911.6 4026.7 4143.2 4261.5

6.5451 6.7671 6.9565 7.1263 7.2826 7.4286 7.5662 7.6970 7.8218 7.9416 8.0569 8.1681 8.2758

◦C

Appendix A

909

TABLE A.5 (Continued )

Thermodynamic Properties of Steam: Superheated Vapor Table P, kPa

T,

◦C

2500 (Tsat = 223.9◦ C) 250 300 350 400 450 500 550 600 650 700 750 800 850 3000 (Tsat = 233.8◦ C) 250 300 350 400 450 500 550 600 650 700 750 800 850 4000 (Tsat = 250.3◦ C) 300 350 400 450 500 550 600 650 700 750 800 850

v, m3 /kg

0.08699 0.09893 0.10975 0.12004 0.13005 0.13987 0.14958 0.15921 0.16876 0.17827 0.18772 0.19714 0.20653

0.07055 0.08116 0.09053 0.09931 0.10779 0.11608 0.12426 0.13234 0.14036 0.14832 0.15624 0.16412 0.17197

0.058835 0.066448 0.073377 0.079959 0.086343 0.092599 0.098764 0.10486 0.11090 0.11690 0.12285 0.12878

u, kJ/kg

2662.2 2762.0 2852.2 2938.9 3024.6 3110.3 3196.6 3284.1 3372.8 3462.9 3554.6 3648.0 3742.9

2643.2 2750.6 2844.3 2932.9 3019.8 3106.3 3193.2 3281.1 3370.2 3460.6 3552.5 3646.0 3741.1

2725.8 2827.6 2920.6 3010.0 3098.2 3186.4 3275.2 3364.9 3455.9 3548.2 3642.1 3737.6

h, kJ/kg

2879.7 3009.3 3126.6 3239.1 3349.7 3459.9 3570.6 3682.1 3794.7 3908.6 4024.0 4140.8 4259.3

2854.9 2994.1 3115.9 3230.9 3343.1 3454.5 3566.0 3678.1 3791.3 3905.6 4021.2 4138.4 4257.1

2961.2 3093.4 3214.1 3329.8 3443.6 3556.8 3670.2 3784.3 3899.5 4015.8 4133.5 4252.7

s, kJ/kg-K

6.4076 6.6446 6.8409 7.0145 7.1730 7.3205 7.4592 7.5906 7.7161 7.8362 7.9518 8.0633 8.1712

6.2855 6.5399 6.7437 6.9213 7.0822 7.2312 7.3709 7.5031 7.6291 7.7497 7.8656 7.9774 8.0855

6.3622 6.5835 6.7699 6.9358 7.0879 7.2298 7.3636 7.4907 7.6121 7.7287 7.8411 7.9496

P, kPa

T,

v, m3 /kg

u, kJ/kg

h, kJ/kg

s, kJ/kg-K

5000 (Tsat = 263.9◦ C) 300 350 400 450 500 550 600 650 700 750 800 850

0.045302 0.051943 0.057792 0.063252 0.068495 0.073603 0.078617 0.083560 0.088447 0.093289 0.098094 0.102867

2698.2 2809.9 2907.7 3000.0 3090.0 3179.5 3269.2 3359.6 3451.1 3543.9 3638.2 3734.0

2924.7 3069.6 3196.6 3316.2 3432.5 3547.5 3662.3 3777.4 3893.4 4010.4 4128.7 4248.3

6.2085 6.4512 6.6474 6.8188 6.9743 7.1184 7.2538 7.3820 7.5044 7.6216 7.7345 7.8435

6000 (Tsat = 275.6◦ C) 300 350 400 450 500 550 600 650 700 750 800 850

0.036146 0.042223 0.047380 0.052104 0.056592 0.060937 0.065185 0.069360 0.073479 0.077552 0.081588 0.085592

2667.1 2790.9 2894.3 2989.7 3081.7 3172.5 3263.1 3354.3 3446.4 3539.6 3634.3 3730.4

2884.0 3044.2 3178.6 3302.3 3421.3 3538.1 3654.2 3770.4 3887.2 4004.9 4123.8 4243.9

6.0669 6.3354 6.5429 6.7202 6.8793 7.0258 7.1627 7.2922 7.4154 7.5333 7.6467 7.7562

7000 (Tsat = 285.8◦ C) 300 350 400 450 500 550 600 650 700 750 800 850

0.029459 0.035234 0.039922 0.044132 0.048087 0.051890 0.055591 0.059218 0.062788 0.066312 0.069798 0.073253

2631.9 2770.6 2880.4 2979.1 3073.2 3165.4 3257.0 3348.9 3441.6 3535.3 3630.3 3726.7

2838.1 3017.2 3159.8 3288.1 3409.8 3528.6 3646.1 3763.4 3881.1 3999.5 4118.9 4239.5

5.9299 6.2301 6.4504 6.6343 6.7971 6.9460 7.0846 7.2153 7.3394 7.4580 7.5720 7.6818

◦C

(continued )

910

Appendix A

TABLE A.5 (Continued )

Thermodynamic Properties of Steam: Superheated Vapor Table P, kPa

T,

◦C

8000 (Tsat =295.0◦ C) 300 350 400 450 500 550 600 650 700 750 800 850 10,000 (Tsat =311.0◦ C) 350 400 450 500 550 600 650 700 750 800 850 15,000 (Tsat = 342.4◦ C) 350 400 450 500 550 600 650 700 750 800 850 20,000 (Tsat = 365.7◦ C) 400 450 500 550 600 650 700 750 800 850 Source:

v, m3 /kg

u, kJ/kg

h, kJ/kg

s, kJ/kg-K

P, kPa

◦C

T,

v, m3 /kg

u, kJ/kg

h, kJ/kg

s, kJ/kg-K

500 550 600 650 700 750 800 850

0.011128 0.012721 0.014126 0.015416 0.016631 0.017790 0.018907 0.019990

2894.0 3024.0 3139.0 3247.0 3351.6 3454.6 3557.2 3659.9

3172.2 3342.1 3492.2 3632.4 3767.4 3899.4 4029.8 4159.6

5.9731 6.1861 6.3632 6.5194 6.6618 6.7941 6.9186 7.0368

500 550 600 650 700 750 800 850

0.008681 0.010166 0.011437 0.012582 0.013647 0.014655 0.015619 0.016549

2833.1 2980.2 3103.9 3217.3 3325.6 3431.4 3536.2 3640.7

3093.6 3285.2 3447.0 3594.7 3735.0 3871.1 4004.8 4137.2

5.8080 6.0484 6.2393 6.4039 6.5519 6.6882 6.8158 6.9364

600 650 700 750 800 850

0.009520 0.010562 0.011521 0.012420 0.013275 0.014095

3068.0 3187.1 3299.2 3408.0 3515.0 3621.4

3401.2 3556.7 3702.5 3842.6 3979.6 4114.7

6.1278 6.3011 6.4548 6.5953 6.7260 6.8491

650 700 750 800 850

0.009053 0.009930 0.010748 0.011521 0.012259

3156.5 3272.6 3384.2 3493.6 3601.8

3518.7 3669.9 3814.2 3954.4 4092.2

6.2077 6.3672 6.5118 6.6456 6.7712

25,000 0.024265 0.029949 0.034311 0.038145 0.041705 0.045103 0.048395 0.051612 0.054770 0.057883 0.060957 0.064000

2592.0 2748.7 2865.8 2968.3 3064.6 3158.2 3250.8 3343.5 3436.8 3531.0 3626.4 3723.1

2786.2 2988.3 3140.3 3273.5 3398.2 3519.0 3638.0 3756.4 3874.9 3994.0 4114.0 4235.1

5.7926 6.1316 6.3665 6.5574 6.7243 6.8757 7.0160 7.1479 7.2729 7.3922 7.5068 7.6170 30,000

0.022422 0.026409 0.029743 0.032760 0.035598 0.038320 0.040964 0.043547 0.046083 0.048581 0.051047

2699.5 2834.8 2945.8 3046.9 3143.6 3238.4 3332.6 3427.0 3522.2 3618.4 3715.8

2923.7 3098.9 3243.2 3374.5 3499.6 3621.6 3742.2 3862.5 3983.0 4104.2 4226.3

5.9448 6.2156 6.4226 6.5982 6.7549 6.8988 7.0332 7.1601 7.2809 7.3965 7.5077 35,000

0.011460 0.015662 0.018452 0.020796 0.022909 0.024884 0.026768 0.028587 0.030356 0.032086 0.033783

2536.0 2742.8 2884.1 3000.2 3105.7 3206.3 3304.8 3402.4 3500.1 3598.3 3697.4

2707.9 2977.8 3160.8 3312.1 3449.3 3579.6 3706.3 3831.2 3955.4 4079.6 4204.2

5.4667 5.8845 6.1472 6.3496 6.5216 6.6753 6.8164 6.9482 7.0727 7.1912 7.3046 40,000

0.009948 0.012707 0.014772 0.016548 0.018162 0.019672 0.021112 0.022499 0.023845 0.025159

2622.8 2812.4 2949.4 3065.8 3173.2 3276.2 3377.2 3477.5 3577.9 3678.7

2821.8 3066.5 3244.8 3396.8 3536.4 3669.7 3799.5 3927.5 4054.8 4181.9

5.5595 5.9110 6.1497 6.3402 6.5049 6.6533 6.7093 6.9186 7.0400 7.1558

Properties obtained from software, STEAMCALC. John Wiley & Sons, New York, 1983.

Appendix A

911

TABLE A.6

Thermodynamic Properties of Refrigerant R-134a: Temperature Table v, m3 /kg; u, kJ/kg; h, kJ/kg; s, kJ/kg-K Specific Volume

Internal Energy

Enthalpy

Entropy

Temp., ◦C T

Press., bar P

Sat. Liquid vf × 103

Sat. Vapor vg

Sat. Liquid uf

Sat. Vapor ug

Sat. Liquid hf

Evap. hfg

Sat. Vapor hg

Sat. Liquid sf

Sat. Vapor sg

−40 −36 −32 −28 −26 −24 −22 −20 −18 −16 −12 −8 −4 0 4 8 12 16 20 24 26 28 30 32 34 36 38 40 42 44 48 52 56 60 70 80 90 100

0.5164 0.6332 0.7704 0.9305 1.0199 1.1160 1.2192 1.3299 1.4483 1.5748 1.8540 2.1704 2.5274 2.9282 3.3765 3.8756 4.4294 5.0416 5.7160 6.4566 6.8530 7.2675 7.7006 8.1528 8.6247 9.1168 9.6298 10.164 10.720 11.299 12.526 13.851 15.278 16.813 21.162 26.324 32.435 39.742

0.7055 0.7113 0.7172 0.7233 0.7265 0.7296 0.7328 0.7361 0.7395 0.7428 0.7498 0.7569 0.7644 0.7721 0.7801 0.7884 0.7971 0.8062 0.8157 0.8257 0.8309 0.8362 0.8417 0.8473 0.8530 0.8590 0.8651 0.8714 0.8780 0.8847 0.8989 0.9142 0.9308 0.9488 1.0027 1.0766 1.1949 1.5443

0.3569 0.2947 0.2451 0.2052 0.1882 0.1728 0.1590 0.1464 0.1350 0.1247 0.1068 0.0919 0.0794 0.0689 0.0600 0.0525 0.0460 0.0405 0.0358 0.0317 0.0298 0.0281 0.0265 0.0250 0.0236 0.0223 0.0210 0.0199 0.0188 0.0177 0.0159 0.0142 0.0127 0.0114 0.0086 0.0064 0.0046 0.0027

−0.04 4.68 9.47 14.31 16.75 19.21 21.68 24.17 26.67 29.18 34.25 39.38 44.56 49.79 55.08 60.43 65.83 71.29 76.80 82.37 85.18 88.00 90.84 93.70 96.58 99.47 102.38 105.30 108.25 111.22 117.22 123.31 129.51 135.82 152.22 169.88 189.82 218.60

204.45 206.73 209.01 211.29 212.43 213.57 214.70 215.84 216.97 218.10 220.36 222.60 224.84 227.06 229.27 231.46 233.63 235.78 237.91 240.01 241.05 242.08 243.10 244.12 245.12 246.11 247.09 248.06 249.02 249.96 251.79 253.55 255.23 256.81 260.15 262.14 261.34 248.49

0.00 4.73 9.52 14.37 16.82 19.29 21.77 24.26 26.77 29.30 34.39 39.54 44.75 50.02 55.35 60.73 66.18 71.69 77.26 82.90 85.75 88.61 91.49 94.39 97.31 100.25 103.21 106.19 109.19 112.22 118.35 124.58 130.93 137.42 154.34 172.71 193.69 224.74

222.88 220.67 218.37 216.01 214.80 213.57 212.32 211.05 209.76 208.45 205.77 203.00 200.15 197.21 194.19 191.07 187.85 184.52 181.09 177.55 175.73 173.89 172.00 170.09 168.14 166.15 164.12 162.05 159.04 157.79 153.33 148.66 143.75 138.57 124.08 106.41 82.63 34.40

222.88 225.40 227.90 230.38 231.62 232.85 234.08 235.31 236.53 237.74 240.15 242.54 244.90 247.23 249.53 251.80 254.03 256.22 258.36 260.45 261.48 262.50 263.50 264.48 265.45 266.40 267.33 268.24 269.14 270.01 271.68 273.24 274.68 275.99 278.43 279.12 276.32 259.13

0.0000 0.0201 0.0401 0.0600 0.0699 0.0798 0.0897 0.0996 0.1094 0.1192 0.1388 0.1583 0.1777 0.1970 0.2162 0.2354 0.2545 0.2735 0.2924 0.3113 0.3208 0.3302 0.3396 0.3490 0.3584 0.3678 0.3772 0.3866 0.3960 0.4054 0.4243 0.4432 0.4622 0.4814 0.5302 0.5814 0.6380 0.7196

0.9560 0.9506 0.9456 0.9411 0.9390 0.9370 0.9351 0.9332 0.9315 0.9298 0.9267 0.9239 0.9213 0.9190 0.9169 0.9150 0.9132 0.9116 0.9302 0.9089 0.9082 0.9076 0.9070 0.9064 0.9058 0.9053 0.9047 0.9041 0.9035 0.9030 0.9017 0.9004 0.8990 0.8973 0.8918 0.8827 0.8655 0.8117

Source:

Properties generated from D. P. Wilson and R. S. Basu, Thermodynamic Properties of a New Stratospherically Safe Working Fluid—Refrigerant 134a. ASHRAE Trans., (94) 2, 2095–2118 (1988).

912

Appendix A

TABLE A.7

Thermodynamic Properties of Refrigerant R-134a: Pressure Table v, m3 /kg; u, kJ/kg; h, kJ/kg; s, kJ/kg-K Specific Volume

Internal Energy

Enthalpy

Entropy

Press., bar P

Temp., ◦C T

Sat. Liquid vf × 103

Sat. Vapor vg

Sat. Liquid uf

Sat. Vapor ug

Sat. Liquid hf

Evap. hfg

Sat. Vapor hg

Sat. Liquid sf

Sat. Vapor sg

0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.4 2.8 3.2 3.6 4.0 5.0 6.0 7.0 8.0 9.0 10.0 12.0 14.0 16.0 18.0 20.0 25.0 30.0

−37.07 −31.21 −26.43 −22.36 −18.80 −15.62 −12.73 −10.09 −5.37 −1.23 2.48 5.84 8.93 15.74 21.58 26.72 31.33 35.53 39.39 46.32 52.43 57.92 62.91 67.49 77.59 86.22

0.7097 0.7184 0.7258 0.7323 0.7381 0.7435 0.7485 0.7532 0.7618 0.7697 0.7770 0.7839 0.7904 0.8056 0.8196 0.8328 0.8454 0.8576 0.8695 0.8928 0.9159 0.9392 0.9631 0.9878 1.0562 1.1416

0.3100 0.2366 0.1917 0.1614 0.1395 0.1229 0.1098 0.0993 0.0834 0.0719 0.0632 0.0564 0.0509 0.0409 0.0341 0.0292 0.0255 0.0226 0.0202 0.0166 0.0140 0.0121 0.0105 0.0093 0.0069 0.0053

3.41 10.41 16.22 21.23 25.66 29.66 33.31 36.69 42.77 48.18 53.06 57.54 61.69 70.93 78.99 86.19 92.75 98.79 104.42 114.69 123.98 132.52 140.49 148.02 165.48 181.88

206.12 209.46 212.18 214.50 216.52 218.32 219.94 221.43 224.07 226.38 228.43 230.28 231.97 235.64 238.74 241.42 243.78 245.88 247.77 251.03 253.74 256.00 257.88 259.41 261.84 262.16

3.46 10.47 16.29 21.32 25.77 29.78 33.45 36.84 42.95 48.39 53.31 57.82 62.00 71.33 79.48 86.78 93.42 99.56 105.29 115.76 125.26 134.02 142.22 149.99 168.12 185.30

221.27 217.92 215.06 212.54 210.27 208.19 206.26 204.46 201.14 198.33 195.35 192.76 190.32 184.74 179.71 175.07 170.73 166.62 162.68 155.23 148.14 141.31 134.60 127.95 111.06 92.71

224.72 228.39 231.35 233.86 236.04 237.97 239.71 241.30 244.09 246.52 248.66 250.58 252.32 256.07 259.19 261.85 264.15 266.18 267.97 270.99 273.40 275.33 276.83 277.94 279.17 278.01

0.0147 0.0440 0.0678 0.0879 0.1055 0.1211 0.1352 0.1481 0.1710 0.1911 0,2089 0.2251 0.2399 0.2723 0.2999 0.3242 0.3459 0.3656 0.3838 0.4164 0.4453 0.4714 0.4954 0.5178 0.5687 0.6156

0.9520 0.9447 0.9395 0.9354 0.9322 0.9295 0.9273 0.9253 0.9222 0.9197 0.9177 0.9166 0.9145 0.9117 0.9097 0.9080 0.9066 0.9054 0.9043 0.9023 0.9003 0.8982 0.8959 0.8934 0.8854 0.8735

Source:

Properties generated from D. P. Wilson and R. S. Basu, Thermodynamic Properties of a New Stratospherically Safe Working Fluid—Refrigerant 134a. ASHRAE Trans., (94) 2, 2095–2118 (1988).

Appendix A

913

TABLE A.8

Thermodynamic Properties of Refrigerant R-134a: Superheated Vapor Table T,◦ C; v, m3 /kg; u, kJ/kg; h, kJ/kg; s, kJ/kg-K T

v

u

h

0.6 bar (0.060 MPa) (Tsat = Sat. −20 −10 0 10 20 30 40 50 60 70 80 90

0.31003 0.33536 0.34992 0.36433 0.37861 0.39279 0.40688 0.42091 0.43487 0.44879 0.46266 0.47650 0.49031

206.12 217.86 224.97 232.24 239.69 247.32 255.12 263.10 271.25 279.58 288.08 296.75 305.58

224.72 237.98 245.96 254.10 262.41 270.89 279.53 288.35 297.34 306.51 315.84 325.34 335.00

s −37.07◦ C) 0.9520 1.0062 1.0371 1.0675 1.0973 1.1267 1.1557 1.1844 1.2126 1.2405 1.2681 1.2954 1.3224

1.4 bars (0.14 MPa) (Tsat = −18.80◦ C) Sat. −10 0 10 20 30 40 50 60 70 80 90

0.13945 0.14549 0.15219 0.15875 0.16520 0.17155 0.17783 0.18404 0.19020 0.19633 0.20241 0.20846

216.52 223.03 230.55 238.21 246.01 253.96 262.06 270.32 278.74 287.32 296.06 304.95

236.04 243.40 251.86 260.43 269.13 277.97 286.96 296.09 305.37 314.80 324.39 334.14

0.9322 0.9606 0.9922 1.0230 1.0532 1.0828 1.1120 1.1407 1.1690 1.1969 1.2244 1.2516

2.0 bars (0.20 MPa) (Tsat = −10.09◦ C) Sat. −10 0 10 20 30 40 50 60 70 80 90

0.09933 0.09938 0.10438 0.10922 0.11394 0.11856 0.12311 0.12758 0.13201 0.13639 0.14073 0.14504

221.43 221.50 229.23 237.05 244.99 253.06 261.26 269.61 278.10 286.74 295.53 304.47

241.30 241.38 250.10 258.89 267.78 276.77 285.88 295.12 304.50 314.02 323.68 333.48

0.9253 0.9256 0.9582 0.9898 1.0206 1.0508 1.0804 1.1094 1.1380 1.1661 1.1939 1.2212

v

u

h

1.0 bar (0.10 MPa) (Tsat = 0.19170 0.19770 0.20686 0.21587 0.22473 0.23349 0.24216 0.25076 0.25930 0.26779 0.27623 0.28464 0.29302

212.18 216.77 224.01 231.41 238.96 246.67 254.54 262.58 270.79 279.16 287.70 296.40 305.27

s −26.43◦ C)

231.35 236.54 244.70 252.99 261.43 270.02 278.76 287.66 296.72 305.94 315.32 324.87 334.57

0.9395 0.9602 0.9918 1.0227 1.0531 1.0829 1.1122 1.1411 1.1696 1.1977 1.2254 1.2528 1.2799

1.8 bars (0.18 MPa) (Tsat = −12.73◦ C) 0.10983 0.11135 0.11678 0.12207 0.12723 0.13230 0.13730 0.14222 0.14710 0.15193 0.15672 0.16148

219.94 222.02 229.67 237.44 245.33 253.36 261.53 269.85 278.31 286.93 295.71 304.63

239.71 242.06 250.69 259.41 268.23 277.17 286.24 295.45 304.79 314.28 323.92 333.70

0.9273 0.9362 0.9684 0.9998 1.0304 1.0604 1.0898 1.1187 1.1472 1.1753 1.2030 1.2303

2.4 bars (0.24 MPa) (Tsat = −5.37◦ C) 0.08343

224.07

244.09

0.9222

0.08574 0.08993 0.09399 0.09794 0.10181 0.10562 0.10937 0.11307 0.11674 0.12037

228.31 236.26 244.30 252.45 260.72 269.12 277.67 286.35 295.18 304.15

248.89 257.84 266.85 275.95 285.16 294.47 303.91 313.49 323.19 333.04

0.9399 0.9721 1.0034 1.0339 1.0637 1.0930 1.1218 1.1501 1.1780 1.2055 (continued )

914

Appendix A

TABLE A.8 (Continued )

Thermodynamic Properties of Refrigerant R-134a: Superheated Vapor Table T,◦ C; v, m3 /kg; u, kJ/kg; h, kJ/kg; s, kJ/kg-K T

v

u

h

2.8 bars (0.28 MPa) (Tsat = Sat. 0 10 20 30 40 50 60 70 80 90 100

0.07193 0.07240 0.07613 0.07972 0.08320 0.08660 0.08992 0.09319 0.09641 0.09960 0.10275 0.10587

226.38 227.37 235.44 243.59 251.83 260.17 268.64 277.23 285.96 294.82 303.83 312.98

246.52 247.64 256.76 265.91 275.12 284.42 293.81 303.32 312.95 322.71 332.60 342.62

s −1.23◦ C) 0.9197 0.9238 0.9566 0.9883 1.0192 1.0494 1.0789 1.1079 1.1364 1.1644 1.1920 1.2193

4.0 bars (0.40 MPa) (Tsat = 8.93◦ C) Sat. 10 20 30 40 50 60 70 80 90 100 110

0.05089 0.05119 0.05397 0.05662 0.05917 0.06164 0.06405 0.06641 0.06873 0.07102 0.07327 0.07550

231.97 232.87 241.37 249.89 258.47 267.13 275.89 284.75 293.73 302.84 312.07 321.44

252.32 253.35 262.96 272.54 282.14 291.79 301.51 311.32 321.23 331.25 341.38 351.64

0.9145 0.9182 0.9515 0.9837 1.0148 1.0452 1.0748 1.1038 1.1322 1.1602 1.1878 1.2149

6.0 bars (0.60 MPa) (Tsat = 21.58◦ C) Sat. 30 40 50 60 70 80 90 100 110 120 130

0.03408 0.03581 0.03774 0.03958 0.04134 0.04304 0.04469 0.04631 0.04790 0.04946 0.05099 0.05251

238.74 246.41 255.45 264.48 273.54 282.66 291.86 301.14 310.53 320.03 329.64 339.93

259.19 267.89 278.09 288.23 298.35 308.48 318.67 328.93 339.27 349.70 360.24 370.88

0.9097 0.9388 0.9719 1.0037 1.0346 1.0645 1.0938 1.1225 1.1505 1.1781 1.2053 1.2320

v

u

h

3.2 bars (0.32 MPa) (Tsat =

s 2.48◦ C)

0.06322

228.43

248.66

0.9177

0.06576 0.06901 0.07214 0.07518 0.07815 0.08106 0.08392 0.08674 0.08953 0.09229

234.61 242.87 251.19 259.61 268.14 276.79 285.56 294.46 303.50 312.68

255.65 264.95 274.28 283.67 293.15 302.72 312.41 322.22 332.15 342.21

0.9427 0.9749 1.0062 1.0367 1.0665 1.0957 1.1243 1.1525 1.1802 1.2076

5.0 bars (0.50 MPa) (Tsat = 15.74◦ C) 0.04086

235.64

256.07

0.9117

0.04188 0.04416 0.04633 0.04842 0.05043 0.05240 0.05432 0.05620 0.05805 0.05988

239.40 248.20 256.99 265.83 274.73 283.72 292.80 302.00 311.31 320.74

260.34 270.28 280.16 290.04 299.95 309.92 319.96 330.10 340.33 350.68

0.9264 0.9597 0.9918 1.0229 1.0531 1.0825 1.1114 1.1397 1.1675 1.1949

7.0 bars (0.70 MPa) (Tsat = 26.72◦ C) 0.02918 0.02979 0.03157 0.03324 0.03482 0.03634 0.03781 0.03924 0.04064 0.04201 0.04335 0.04468

241.42 244.51 253.83 263.08 272.31 281.57 290.88 300.27 309.74 319.31 328.98 338.76

261.85 265.37 275.93 286.35 296.69 307.01 317.35 327.74 338.19 348.71 359.33 370.04

0.9080 0.9197 0.9539 0.9867 1.0182 1.0487 1.0784 1.1074 1.1358 1.1637 1.1910 1.2179

Appendix A

915

TABLE A.8 (Continued )

Thermodynamic Properties of Refrigerant R-134a: Superheated Vapor Table T,◦ C; v, m3 /kg; u, kJ/kg; h, kJ/kg; s, kJ/kg-K T Sat. 40 50 60 70 80 90 100 110 120 130 140

v

u

h

s

8.0 bars (0.80 MPa) (Tsat = 31.33◦ C) 0.02547 243.78 264.15 0.9066 0.02691 252.13 273.66 0.9374 0.02846 261.62 284.39 0.9711 0.02992 271.04 294.98 1.0034 0.03131 280.45 305.50 1.0345 0.03264 289.89 316.00 1.0647 0.03393 299.37 326.52 1.0940 0.03519 308.93 337.08 1.1227 0.03642 318.57 347.71 1.1508 0.03762 328.31 358.40 1.1784 0.03881 338.14 369.19 1.2055 0.03997 348.09 380.07 1.2321 10.0 bars (1.00 MPa) (Tsat = 39.39◦ C)

Sat. 40 50 60 70 80 90 100 110 120 130 140

0.02020 0.02029 0.02171 0.02301 0.02423 0.02538 0.02649 0.02755 0.02858 0.02959 0.03058 0.03154

247.77 248.39 258.48 268.35 278.11 287.82 297.53 307.27 317.06 326.93 336.88 346.92

267.97 268.68 280.19 291.36 302.34 313.20 324.01 334.82 345.65 356.52 367.46 378.46

0.9043 0.9066 0.9428 0.9768 1.0093 1.0405 1.0707 1.1000 1.1286 1.1567 1.1841 1.2111

14.0 bars (1.40 MPa) (Tsat = 52.43◦ C)

v

u

h

s

9.0 bars (0.90 MPa) (Tsat = 35.53◦ C) 0.02255 245.88 266.18 0.9054 0.02325 250.32 271.25 0.9217 0.02472 260.09 282.34 0.9566 0.02609 269.72 293.21 0.9897 0.02738 279.30 303.94 1.0214 0.02861 288.87 314.62 1.0521 0.02980 298.46 325.28’ 1.0819 0.03095 308.11 335.96 1.1109 0.03207 317.82 346.68 1.1392 0.03316 327.62 357.47 1.1670 0.03423 337.52 368.33 1.1943 0.03529 347.51 379.27 1.2211 12.0 bars (1.20 MPa) (Tsat = 46.32◦ C) 0.01663

251.03

270.99

0.9023

0.01712 0.01835 0.01947 0.02051 0.02150 0.02244 0.02335 0.02423 0.02508 0.02592

254.98 265.42 275.59 285.62 295.59 305.54 315.50 325.51 335.58 345.73

275.52 287.44 298.96 310.24 321.39 332.47 343.52 354.58 365.68 376.83

0.9164 0.9527 0.9868 1.0192 1.0503 1.0804 1.1096 1.1381 1.1660 1.1933

16.0 bars (1.60 MPa) (Tsat = 57.92◦ C)

Sat. 0.01405 253.74 273.40 0.9003 0.01208 256.00 275.33 0.8982 60 0.01495 262.17 283.10 0.9297 0.01233 258.48 278.20 0.9069 70 0.01603 272.87 295.31 0.9658 0.01340 269.89 291.33 0.9457 80 0.01701 283.29 307.10 0.9997 0.01435 280.78 303.74 0.9813 90 0.01792 293.55 318.63 1.0319 0.01521 291.39 315.72 1.0148 100 0.01878 303.73 330.02 1.0628 0.01601 301.84 327.46 1.0467 110 0.01960 313.88 341.32 1.0927 0.01677 312.20 339.04 1.0773 120 0.02039 324.05 352.59 1.1218 0.01750 322.53 350.53 1.1069 130 0.02115 334.25 363.86 1.1501 0.01820 332.87 361.99 1.1357 140 0.02189 344.50 375.15 1.1777 0.01887 343.24 373.44 1.1638 150 0.02262 354.82 386.49 1.2048 0.01953 353.66 384.91 1.1912 160 0.02333 365.22 397.89 1.2315 0.02017 364.15 369.43 1.2181 Source: Properties generated from D. P. Wilson and R. S. Basu, Thermodynamic Properties of a New Stratospherically Safe Working Fluid—Refrigerant 134a. ASHRAE Trans., (94) 2, 2095–2118 (1988).

916

Appendix A

TABLE A.9

Critical Constants

Gas Acetylene Air Ammonia Argon Benzene Butane Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Methane Nitrogen Octane Oxygen Propane Propylene Refrigerant 12 Refrigerant 134a Water

UFM|484942|1435612135

Source:

Critical Temperature, K

Critical Pressure, bar

Critical Compressibility, Zc

Molecular Weight, M

309 133 406 151 563 425 304 133 305 283 5.2 33.2 191 126 569 154 370 365 385 374 647.3

62.8 37.7 112.8 48.6 49.3 38.0 73.9 35.0 48.8 63.8 2.3 13.0 46.4 33.9 24.9 50.5 42.7 46.2 41.2 40.7 220.9

0.274 0.284 0.242 0.290 0.274 0.274 0.276 0.294 0.285 0.270 0.300 0.304 0.290 0.291 0.258 0.290 0.276 0.276 0.278 0.260 0.233

26.04 28.97 17.04 39.94 78.11 58.12 44.01 28.01 30.01 28.05 4.003 2.016 16.04 28.01 114.22 32.00 44.09 42.08 120.92 102.03 18.02

L. C. Nelson and E. F. Obert, Generalized Compressibility Charts. Chem. Eng. 61, 203 (1954).

Appendix A

917

TABLE A.10

Constants for the van der Waals Equation of Statea Gas Acetylene Air Ammonia Benzene Butane Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Methane Nitrogen Oxygen Propane Refrigerant 134a Water a

ar bar-(m3 /kg mol))2

br m3 /kg mol

4.410 1.368 4.223 18.63 13.860 3.647 1.474 5.575 4.563 0.0341 0.2476 2.293 1.366 1.369 9.349 10.05 5.531

0.0510 0.0367 0.0373 0.1181 0.1162 0.0428 0.0395 0.0650 0.0574 0.0234 0.0265 0.0428 0.0386 0.0317 0.0901 0.0957 0.0305

Determined from the generalized behavior of gases represented by the derivatives ∂ P/∂v¯ = 0 and ∂2 P/∂v¯ 2 = 0.

TABLE A.11

Constants for the Redlich-Kwong Equation of Statea Gas Carbon dioxide Carbon monoxide Methane Nitrogen Oxygen Propane Refrigerant 134a Water a

ar bar-(m3 /kg mol)2 K1/2

br m3 /kg mol

64.64 17.26 32.19 15.59 17.38 183.070 197.1 142.64

0.02969 0.02743 0.02969 0.002681 0.02199 0.06269 0.06634 0.02110

Determined from the generalized behavior of gases represented by the derivatives ∂ P/∂v¯ = 0 and ∂2 P/∂v¯ 2 = 0.

918

Appendix A

TABLE A.12

Thermodynamic Properties of Air T, K

Pr

vr

u, kJ/kg

h, kJ/kg

s◦ , kJ/kg-K

230 240 250 260 270 280 285 290 295 300 305 310 315 320 325 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 600 610

0.5477 0.6355 0.7329 0.8405 0.9590 1.0889 1.1584 1.2311 1.3068 1.3860 1.4686 1.5546 1.6442 1.7375 1.8345 1.9352 2.149 2.379 2.626 2.892 3.176 3.481 3.806 4.153 4.522 4.915 5.332 5.775 6.245 6.742 7.268 7.824 8.411 9.031 9.684 10.37 11.10 11.86 12.66 13.50 14.38 15.31 16.28 17.30

1205.0 1084.0 979.0 887.8 808.0 738.0 706.1 676.1 647.9 621.2 596.0 572.3 549.8 528.6 508.4 489.4 454.1 422.2 393.4 367.2 343.4 321.5 301.6 283.3 266.6 251.1 236.8 223.6 211.4 200.1 189.5 179.7 170.6 162.1 154.1 146.7 139.7 133.1 127.0 121.2 115.7 110.6 105.8 101.2

164.00 171.13 178.28 185.45 192.60 199.75 203.33 206.91 210.49 214.07 217.67 221.25 224.85 228.42 232.02 235.61 242.82 250.02 257.24 264.46 271.69 278.93 286.16 293.43 300.69 307.99 315.30 322.82 329.97 337.32 344.70 352.08 359.49 366.92 374.36 381.84 389.34 396.86 404.42 411.97 419.55 427.15 434.78 442.42

230.02 240.02 250.05 260.09 270.11 280.13 285.14 290.16 295.17 300.19 305.22 310.24 315.27 320.29 325.31 330.34 340.42 350.49 360.58 370.67 380.77 390.88 400.98 411.32 421.26 431.43 441.61 451.80 462.02 472.24 482.49 492.74 503.02 513.32 523.63 533.98 544.35 555.74 565.17 575.59 586.04 596.57 607.02 617.53

1.4356 1.4782 1.1925 1.5585 1.5963 1.6328 1.6506 1.6680 1.6852 1.7020 1.7186 1.7350 1.7511 1.7669 1.7825 1.7978 1.8279 1.8571 1.8854 1.9131 1.9400 1.9663 1.9919 2.0170 2.0414 2.0653 2.0887 2.1161 2.1341 2.1560 2.1776 2.1988 2.2195 2.2399 2.2600 2.2797 2.2991 2.3181 2.3369 2.3553 2.3735 2.3914 2.4090 2.4264

Appendix A

919

TABLE A.12 (Continued )

Thermodynamic Properties of Air T, K

Pr

vr

u, kJ/kg

h, kJ/kg

s◦ , kJ/kg-K

620 630 640 650 660 670 680 690 700 720 740 760 780 800 820 840 860 880 900 920 940 960 980 1000 1020 1040 1060 1080 1100 1120 1140 1160 1180 1200 1220 1240 1260 1280 1300 1320 1340 1360 1380 1400 1420

18.36 19.84 20.64 21.86 23.13 24.46 25.85 27.29 28.80 32.02 35.50 39.27 43.35 47.75 52.59 57.60 63.09 68.98 75.29 82.05 89.28 97.00 105.2 114.0 123.4 133.3 143.9 155.2 167.1 179.7 193.1 207.2 222.2 238.0 254.7 272.2 290.8 310.4 330.9 352.5 375.3 399.1 424.2 450.5 478.0

96.92 92.84 88.99 85.34 81.89 78.61 75.50 72.56 69.76 64.53 59.82 55.54 51.64 48.08 44.84 41.85 39.12 36.61 34.31 32.18 30.22 28.40 26.73 25.17 23.72 22.39 21.14 19.98 18.896 17.886 16.946 16.064 15.241 14.470 13.747 13.069 12.435 11.835 11.275 10.747 10.247 9.780 9.337 8.919 8.526

450.09 457.78 465.50 473.25 481.01 488.81 496.62 504.45 512.33 528.14 544.02 560.01 576.12 592.30 608.59 624.95 641.40 657.95 674.58 691.28 708.08 725.02 741.98 758.94 776.10 793.36 810.62 827.88 845.33 862.79 880.35 897.91 915.57 933.33 951.09 968.95 986.90 1004.76 1022.82 1040.88 1058.94 1077.10 1095.26 1113.52 1131.77

628.07 638.63 649.22 659.84 670.47 681.14 691.82 702.51 713.27 734.82 756.44 778.18 800.03 821.95 843.98 866.08 888.27 910.56 932.93 955.38 977.92 1000.55 1023.55 1046.04 1068.89 1091.85 1114.86 1137.89 1161.07 1184.28 1207.57 1230.92 1254.34 1277.79 1301.31 1324.93 1348.55 1372.24 1395.97 1429.76 1443.60 1467.49 1491.44 1515.42 1539.44

2.4436 2.4605 2.4772 2.4936 2.5099 2.5259 2.5418 2.5573 2.5728 2.6032 2.6328 2.6618 2.6901 2.7179 2.7450 2.7717 2.7978 2.8234 2.8459 2.8732 2.8975 2.9213 2.9497 2.9677 2.9903 3.0126 3.0345 3.0561 3.0773 3.0983 3.1188 3.1392 3.1592 3.1789 3.1983 3.2175 3.2364 3.2551 3.2735 3.2916 3.3096 3.3272 3.3447 3.3620 3.3790 (continued )

920

Appendix A

TABLE A.12 (Continued )

Thermodynamic Properties of Air T, K

Pr

vr

u, kJ/kg

h, kJ/kg

s◦ , kJ/kg-K

1440 1460 1480 1500 1520 1540 1560 1580 1600 1620 1640 1660 1680 1700 1750 1800 1850 1900 1950 2000 2050 2100 2150 2200 2250

506.9 537.1 568.8 601.9 636.5 672.8 710.5 750.0 791.2 834.1 878.9 925.6 974.6 1025 1161 1310 1475 1655 1852 2068 2303 2559 2837 3138 3464

8.153 7.801 7.468 7.152 6.854 6.569 6.301 6.046 5.804 5.574 5.355 5.147 4.949 4.761 4.328 3.944 3.601 3.295 3.022 2.776 2.555 2.356 2.175 2.012 1.864

1150.13 1168.49 1186.95 1205.41 1223.87 1242.43 1260.99 1279.65 1298.30 1316.96 1335.72 1354.48 1373.24 1392.7 1439.8 1487.2 1534.9 1582.6 1630.6 1678.7 1726.8 1775.3 1823.8 1872.4 1921.3

1563.51 1587.63 1611.79 1635.97 1660.23 1684.51 1708.82 1733.17 1757.57 1782.00 1806.76 1839.96 1855.50 1880.1 1941.6 2003.2 2065.3 2127.4 2189.7 2251.1 2314.6 2377.7 2440.3 2503.2 2566.4

3.3959 3.4125 3.4289 3.4452 3.4612 3.4771 3.4928 3.5083 3.5236 3.5388 3.5538 3.5687 3.5834 3.5979 3.6336 3.6684 3.7013 3.7354 3.7667 3.7994 3.8303 3.8605 3.8901 3.9191 3.9474

Source: J. H. Keenan and J. Kaye, Gas Tables. John Wiley & Sons, New York, 1945.

Appendix A

921

TABLE A.13

Thermal Properties of Air at Atmospheric Pressure T, K 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 900 1000 1100 1200 1300 1400 1500 1600 1800 2000

ρ,

kg/m3

3.6010 2.3675 1.7684 1.4128 1.1774 0.9980 0.8826 0.7823 0.7048 0.6423 0.5879 0.5430 0.5030 0.4709 0.4405 0.3925 0.3524 0.3204 0.2947 0.2707 0.2515 0.2355 0.2211 0.1970 0.1762

cp , kJ/kg-K

μ, kg/m-s ×105

v, m2 /s ×l06

1.0266 1.0099 1.0061 1.0053 1.0057 1.0090 1.0140 1.0207 1.0295 1.0392 1.0551 1.0635 1.0752 1.0856 1.0978 1.1212 1.1417 1.160 1.179 1.197 1.214 1.230 1.248 1.287 1.338

0.6924 1.0283 1.3289 1.5990 1.8462 2.075 2.286 2.484 2.671 2.848 3.018 3.177 3.332 3.481 3.625 3.899 4.152 4.440 4.690 4.930 5.170 5.400 5.630 6.070 6.50

1.923 4.343 7.490 11.31 15.69 20.76 25.90 31.71 37.90 44.34 51.34 58.51 66.25 73.91 82.29 99.30 117.8 138.6 159.1 182.1 205.5 229.1 254.5 308.1 369.0

k, W/m-K

α,m2 /s ×104

Pr

0.0092 0.0137 0.0181 0.0228 0.0262 0.0300 0.0337 0.0371 0.0404 0.0436 0.0466 0.0495 0.0523 0.0551 0.0578 0.0629 0.0675 0.0732 0.0782 0.0837 0.0891 0.0946 0.1000 0.1110 0.1240

0.0250 0.0575 0.3017 0.1568 0.2216 0.2983 0.3760 0.4220 0.5564 0.6532 0.7512 0.8578 0.9672 1.0774 1.1951 1.4271 1.6679 1.969 2.251 2.583 2.920 3.262 3.609 4.379 5.260

0.770 0.753 0.739 0.722 0.708 0.697 0.689 0.683 0.680 0.680 0.680 0.682 0.684 0.686 0.689 0.696 0.702 0.704 0.707 0.705 0.705 0.705 0.705 0.704 0.700

Source: U.S. National Bureau of Standards, Tables of Thermal Properties of Gases, Circular 564, 1955.

TABLE A.14

Physical Properties of Water at Atmospheric Pressure T,◦ C 0 10 20 30 40 50 60 70 80 90 100 120 140 160 180 200 220 240 260 280 300

ρ, kg/m3

c, kJ/kg-K

μ, kg/m-s ×103

v, m2 /s ×106

k, W/m-K

α, m2 /s ×106

Pr

β, K−1 ×103

1000.0 1000.0 998.0 996.0 992.6 988.1 983.3 977.8 971.8 965.4 958.6 943.4 926.4 907.8 887.3 865.0 840.8 813.7 783.7 750.8 712.3

4.194 4.202 4.190 4.179 4.177 4.178 4.183 4.187 4.197 4.206 4.219 4.251 4.294 4.350 4.420 4.506 4.616 4.761 4.955 5.224 5.610

1.790 1.310 1.010 0.803 0.656 0.536 0.475 0.408 0.359 0.318 0.283 0.229 0.196 0.174 0.154 0.139 0.127 0.116 0.107 0.098 0.091

1.790 1.310 1.012 0.8062 0.6609 0.5435 0.4831 0.4173 0.3694 0.3294 0.2952 0.2427 0.2116 0.1917 0.1736 0.1607 0.1511 0.1426 0.1365 0.1305 0.1278

0.566 0.585 0.602 0.619 0.633 0.644 0.654 0.664 0.671 0.676 0.682 0.085 0.685 0.681 0.676 0.667 0.654 0.638 0.616 0.581 0.530

0.1350 0.1392 0.1440 0.1487 0.1527 0.1560 0.1590 0.1622 0.1645 0.1665 0.1686 0.1708 0.1722 0.1725 0.1724 0.1711 0.1685 0.1647 0.1586 0.1481 0.1326

13.26 9.41 7.03 5.42 4.33 3.48 3.04 2.573 2.245 1.979 1.751 1.421 1.229 1.111 1.007 0.939 0.896 0.866 0.861 0.881 0.963

0.000 0.100 0.200 0.290 0.380 0.460 0.530 0.590 0.640 0.690 0.740 0.860 0.960 1.080 1.200 1.380 1.540 1.700 2.140 2.390 2.880

Source: J. P. Todd and H. B. Ellis, Applied Heat Transfer. Harper & Row, New York, 1982.

922

Appendix A

TABLE A.15

Physical Properties of Gases at Atmospheric Pressure ρ, kg/m3

cp , kJ/kg-K

k, W/m-K

μ, kg/m-s ×106

ν, m2 /s ×106

Pr

200 300 400 500 600 700 800

1.7108 1.1421 0.8538 0.6824 0.5687 0.4934 0.4277

1.0429 1.0408 1.0459 1.0555 1.0756 1.0969 1.1225

0.0182 0.0262 0.0334 0.0398 0.0458 0.0512 0.0561

12.95 17.84 21.96 25.70 29.11 32.13 34.84

7.57 15.63 25.74 27.66 51.19 65.13 81.46

0.747 0.713 0.691 0.684 0.686 0.691 0.700

O2 200 250 300 350 400 450 500 550

1.9959 1.5618 1.3007 1.1133 0.9755 0.8682 0.7801 0.7096

0.9131 0.9157 0.9203 0.9291 0.9420 0.9567 0.9722 0.9881

0.0182 0.0226 0.0268 0.0307 0.0346 0.0383 0.0417 0.0452

14.85 17.87 20.63 23.16 25.54 27.77 29.91 31.07

7.593 11.45 15.86 20.80 26.18 31.99 38.34 45.05

0.745 0.725 0.709 0.702 0.695 0.684 0.697 0.700

CO2 250 300 350 400 450 500 550 600

2.1657 1.7973 1.5362 0.3424 1.1918 1.0732 0.9739 0.8938

0.804 0.871 0.900 0.942 0.980 1.013 1.047 1.076

0.0129 0.0166 0.0205 0.0246 0.0290 0.0335 0.0382 0.0483

12.59 14.96 17.21 19.32 21.34 23.26 25.08 26.83

5.813 8.321 11.19 14.39 17.90 21.67 25.74 30.02

0.793 0.770 0.747 0.738 0.721 0.702 0.685 0.686

CO 300 350 400 450 500 550 600

1.139 0.974 0.854 0.758 0.682 0.620 0.569

1.042 1.043 1.048 1.055 1.064 1.076 1.088

0.0253 0.0288 0.0323 0.0356 0.0386 0.0416 0.0444

17.84 20.09 22.19 24.18 26.06 27.89 29.60

15.67 20.62 25.99 31.88 38.19 44.97 52.06

0.737 0.728 0.722 0.718 0.718 0.721 0.724

T, K N2

Appendix A

923

TABLE A.15 (Continued )

Physical Properties of Gases at Atmospheric Pressure ρ, kg/m3

cp , kJ/kg-K

k, W/m-K

μ, kg/m-s ×106

ν, m2 /s ×106

Pr

150 200 250 300 350 400 450 500 600 700 800 900

0.1637 0.1227 0.0982 0.0819 0.0702 0.0614 0.0546 0.0492 0.0409 0.0349 0.0306 0.0272

12.602 13.540 14.069 14.314 14.436 14.491 14.499 14.507 14.537 14.574 14.675 14.821

0.0981 0.1282 0.1561 0.182 0.208 0.228 0.251 0.272 0.315 0.351 0.384 0.412

5.595 6.813 7.919 8.963 9.954 10.864 11.779 12.636 14.285 15.89 17.40 18.78

34.18 55.53 80.64 109.5 141.9 177.1 215.6 257.0 349.7 455.1 569.0 690.0

0.718 0.719 0.713 0.706 0.697 0.690 0.682 0.675 0.664 0.659 0.664 0.676

He ` I44 200 255 368 477 589 700 800

0.3379 0.2435 0.1906 0.1328 0.1020 0.0828 0.0703 0.0602

5.200 5.200 5.200 5.200 5.200 5.200 5.200 5.200

0.0928 0.1177 0.1357 0.1691 0.197 0.225 0.251 0.275

12.55 16.66 18.17 23.05 27.50 31.13 34.75 38.17

37.11 64.38 95.50 173.6 269.3 375.8 494.2 634.1

0.700 0.694 0.700 0.710 0.720 0.720 0.720 0.720

T, K H2

Steam 400 450 500 550 600 650 700 750 800 850 Source:

0.5542 2.014 0.0261 13.44 24.2 1.040 0.4902 1.980 0.0299 15.25 31.1 1.010 0.4405 1.985 0.0339 17.04 38.6 0.996 0.4005 1.997 0.0379 18.84 47.0 0.991 0.3652 2.026 0.0422 20.67 56.6 0.986 0.3380 2.056 0.0464 22.47 66.4 0.995 0.3140 2.085 0.0505 24.26 77.2 1.000 0.2931 2.119 0.0549 26.04 88.8 1.005 0.2739 2.152 0.0592 27.86 102.0 1.010 0.2579 2.186 0.0637 29.69 115.2 1.019 Converted to SI units from E. R. G. Eckert and R. M. Drake, Heat and Mass Transfer, 2nd ed. McGraw-Hill, New York, 1959.

924

Appendix A TABLE A.16

Physical Properties of Liquids at Atmospheric Pressure

T, K

ρ, kg/m3

c, kJ/kg-K

k, W/m-K

μ, kg/m-s ×102

ν, m2 /s ×106

Pr

β, K−1 ×103

2.261 2.298 2.367 2.427 2.490 2.564

0.282 0.284 0.286 0.286 0.286 0.287

1060 534 185 79.9 35.2 21.0

8310 4200 1460 634 281 168

85,000 43,200 15,300 6780 3060 1870

0.47 0.47 0.48 0.48 0.49 0.50

2.294 2.323 2.368 2.415 2.505 2.549 2.592 2.637 2.682

0.242 0.244 0.248 0.252 0.255 0.258 0.260 0.261 0.261

6.51 4.20 2.47 1.57 1.07 0.757 0.561 0.431 0.342

57.6 37.3 22.1 14.1 9.65 6.91 5.15 3.98 3.17

617 400 236 151 103 73.5 55.0 42.8 34.6

0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65

1.827 1.868 1.909 1.951 1.993 2.035 2.076 2.118

0.144 0.145 0.145 0.145 0.143 0.141 0.139 0.138

217.0 99.9 48.6 25.3 14.1 8.36 5.31 3.56

2430 1120 550 288 161 96.6 61.7 41.7

27,500 12,900 6400 3400 1965 1205 793 596

0.70 0.70 0.70 0.70 0.70 0.70 0.70 0.70

0.147 0.145 0.144 0.140 0.138 0.137 0.135 0.133 0.132

384.8 79.94 21.03 7.25 3.20 1.71 1.03 0.65 0.45

4280 900 240 83.9 37.5 20.3 12.4 8.0 5.6

47,100 10,400 2870 1050 490 276 175 116 84

Glycerin 273 280 290 300 310 320

1276.0 1271.9 1265.8 1259.9 1253.9 1247.2

Ethylene Glycol 273 280 290 300 310 320 330 340 350

1130.8 1125.8 1118.8 1114.4 1103.7 1096.2 1089.5 1083.8 1079.0

Engine Oil (unused) 280 290 300 310 320 330 340 350

895.3 890.0 884.1 877.9 871.8 865.8 859.9 853.9

Lubricating Oil (≈ SAE 50) 273 293 313 333 353 373 393 413 433

899.12 888.23 876.05 864.04 852.02 840.01 828.96 816.94 805.89

1.796 1.880 1.964 2.047 2.131 2.219 2.307 2.395 2.483

CO2 223 703.69 4.463 0.547 3.06 0.435 2.60 233 691.68 4.467 0.547 2.81 0.406 2.28 243 679.24 4.476 0.549 2.63 0.387 2.15 253 666.69 4.509 0.547 2.54 0.381 2.09 263 653.55 4.564 0.543 2.47 0.378 2.07 273 640.10 4.635 0.540 2.39 0.373 2.05 283 626.16 4.714 0.531 2.30 0.368 2.04 293 611.75 4.798 0.521 2.20 0.359 2.02 303 596.37 4.890 0.507 2.07 0.349 2,01 313 580.99 4.999 0.493 1.98 0.340 2.00 323 564.33 5.116 0.476 1.86 0.330 1.99 Source: Converted to SI units from E. R. G. Eckert and R. M. Drake, Heat and Mass Transfer, 2nd ed., McGraw-Hiil, New York, 1959.

Appendix A

925

TABLE A.17

Thermal Properties of Metals at 20◦ C Metal Aluminum Pure 1% Cu 22% Si 1% Mn Brass, 70% Cu, 30% Zn Copper Iron Wrought iron Lead Magnesium Molybdenum Nickel Silver Steel 0.5% C 1.0% C 1.0% Cr 5.0% Cr Tin Tungsten Zinc Source:

k, W/m-K

p, kg/m3

c, kJ/kg-K

α, m2 /s ×105

20◦ C

100◦ C

200◦ C

300◦ C

2,707 2,659 2,627 2,707 8,522 8,954 7,897 7,849 11,373 1,746 10,220 8,906 10,525

0.896 0.867 0.854 0.892 0.385 0.383 0.452 0.460 0.130 1.013 0.251 0.446 0.234

8.418 5.933 7.172 7.311 3.412 11.234 2.034 1.626 2.343 9.708 3.605 2.266 16.563

204 137 161 177 111 386 73 59 35 171 66 90 407

206 144 168 189 128 379 67 57 33.4 168 62 83 415

215 152 175 204 144 374 62 52 31.5 163 74 73 374

228 161 178 147 369 55 48 29.8 157 83 64 362

7,833 7,801 7,865 7,833 7,304 19,350 7,144

0.465 0.473 0.460 0.460 0.227 0.134 0.384

1.474 1.172 1.665 1.110 3.884 6.271 4.106

54 43 61 40 64 163 112

52 43 55 38 59 151 109

48 42 52 36 57 142 106

45 40 47 36 55 133 100

Converted to SI units from E. R. G. Eckert and R. M. Drake, Heat and Mass Transfer, 2nd ed. McGraw-Hill, New York, 1959.

TABLE A.18

Thermal Properties of Nonmetals Material Asbestos Bakelite Brick Common Masonry Clay Coal, anthracite Concrete, dry Corkboard Glass, window Glass wool Ice Paper Plexiglas Rock wool Rubber, hard Soil, dry Teflon Wood Oak Pine

α × 106 m2 /s

T,◦ C

ρ, kg/m3

c, J/kg-K

20 30

383 1200

816 1600

0.113 0.23

0.036 0.012

20 20 20 20 20 20 20 20 0 30 20 20 20 20 30

1800 1760 1545 1370 2300 150 2700 200 910 900 1200 160 2250 1500 2300

840 837 880 1260 837 1880 840 670 1930 1200 1500 1200 2009 1842 1050

0.38–0.52 0.638 1.26 0.238 1.80 0.042 0.78 0.040 2.22 0.12 0.20 0.040 0.163 0.35 0.35

0.028–0.034 0.0406 0.101 0.013–0.015 0.094 0.015–0.044 0.034 0.028 0.126 0.011 0.0164 0.011 0.0162 0.138 0.112–0.119

20 20

609-801 116–421

2390 2720

0.17–0.21 0.15

0.0111–0.1021 0.0124

k, W/m-K

Source: Converted to SI units from E. R. G. Eckert and R. M. Drake, Heat and Mass Transfer, 2nd ed. McGraw-Hill, New York, 1959.

926

Appendix A

TABLE A.19

Normal Total Emissivities of Several Metals Material Aluminum Polished 24ST Brass, polished Chromium, polished Copper, polished Gold, polished Iron Roughly polished Cast, polished Wrought, polished Lead, unoxidized Magnesium oxide Mercury Molybdenum, polished Nickel, polished Silver, polished Steel Stainless, polished Type 301 Type 316 Type 347 Tin, bright Tungsten, polished Zinc Commercial, polished Galvanized sheet

T,◦ C

ε

100 232–504 100 100 100 227–627

0.095 0.16–0.20 0.06 0.075 0.052 0.018–0.35

100 200 38–250 127–227 277–827 0–100 100 100 100

0.17 0.21 0.28 0.057–0.075 0.55–0.20 0.09–0.12 0.071 0.072 0.052

100 24 24 24 25 100

0.074 0.21 0.28 0.39 0.43 0.066

227–327 100

0.045–0.053 21

Source: From a compilation by H. C. Hottel in W. H. McAdams, Heat Transmission, 3rd ed. McGrawHill, New York, 1954.

Appendix A

927

TABLE A.20

Standard Pipe Sizes Nominal Size, in. 1/2 1 2 3 4 5 6 8 10 12

Schedule Number 40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80

Outside Diameter, cm

Outside Surface m2 /m

Inside Diameter, cm

2.134

0.06704

3.340

0.1049

6.033

0.1895

8.890

0.2793

11.43

0.3591

14.13

0.4439

16.83

0.5287

21.91

0.6683

27.31

0.8580

32.29

0.10176

1.580 1.387 2.664 2.431 5.250 4.925 7.793 7.366 10.226 9.718 12.82 12.23 15.41 14.63 20.27 19.37 25.45 24.29 30.32 28.90

Source: NAVCO Piping Catalog.

928

Appendix A

TABLE A.21

Standard Tubing Sizes Nominal Size, in. 5/8

1/2

3/4

1

1 1/4

1 1/2

2

BWG 18 16 14 18 16 14 18 16 14 18 16 14 12 10 18 16 14 12 10 13 12 11 10 8 13 12 11 10 8

Outside Diameter, cm

Outside Surface, m2 /m

Inside Diameter, cm

1.270

0.03990

1.588

0.04989

1.905

0.05985

2.540

0.07980

3.175

0.09975

3.810

0.1197

5.080

0.1596

1.021 0.940 0.848 1.339 1.257 1.166 1.656 1.575 1.483 2.291 2.210 2.118 1.986 1.589 2.926 2.845 2.753 2.621 2.494 3.327 3.256 3.200 3.129 2.972 4.597 4.526 4.470 4.399 4.242

Source: NAVCO Piping Catalog.

Appendix A

929

TABLE A.22

Properties of the Standard Atmosphere Altitude, h meters

T, K K

P Pa

ρ kg/m3

μ × 105 kg/m-s

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000 7500 8000 8500 9000 9500 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 25,000

288.2 284.9 281.7 278.4 275.2 271.9 268.7 265.4 262.2 258.9 255.7 252.4 249.2 245.9 242.7 239.5 236.2 233.0 229.7 226.5 223.3 216.8 216.7 216.7 216.7 216.7 216.7 216.7 216.7 216.7 216.7 221.5

1.0133 × 105 0.95461 × 105 0.89876 × 105 0.84560 × 105 0.79501 × 105 0.74692 × 105 0.70121 × 105 0.65780 × 105 0.61660 × 105 0.57753 × 105 0.54048 × 105 050539 × 105 0.47218 × 105 0.44075 × 105 0.41105 × 105 0.38300 × 105 0.35652 × 105 0.33154 × 105 0.30801 × 105 0.28585 × 105 0.26500 × 105 0.22700 × 105 0.19399 × 105 0.16580 × 105 0.14170 × 105 0.12112 × 105 0.10353 × 105 8.8597 × 103 5.5293 × 103 2.5492 × 103 5.5293 × 103 2.5492 × 103

1.225 1.167 1.111 1.058 1.007 0.9570 0.9093 0.8634 0.8194 0.7770 0.7364 0.6975 0.6601 0.6243 0.5900 0.5572 0.5258 0.4958 0.4671 0.4397 0.4135 0.3648 0.3119 0.2666 0.2279 0.1948 0.1665 0.1423 0.12I7 0.1040 0.08891 0.04008

1.789 1.774 1.758 1.742 1.726 1.710 1.694 1.678 1.661 1.645 1.628 1.612 1.595 1.578 1.561 1.544 1.527 1.510 1.493 1.475 1.458 1.422 1.422 1.422 1.422 1.422 1.422 1.422 1.422 1.422 1.422 1.448

Source: Extracted from NACA-TN-1428

930

Appendix A

τ, 5.00

1.00

6.00

5.00

4.00

3.50

3.00

1.60

2.60 .40 2 2.20 .00 2

1.80

0.

1.40

1.60 0 1.5 1.40

1.30

1.30

0. 85

Compressibility Factor, a = Pv RT

50

0.80

1.10

0

Note

1.00

0.9

Z 0.95

0 0.7

0.90 0.00

???

0.30 0.0

0.1

0.2

0.3

0.6 0.4 0.5 Reduced Pressure, Pr

τ, 1

.0

0

0.85

5

0.40

v Rτo/Por

Chart No. 1

2.00 1.80 1.40 1.20 1.10 1.05 1.00 0.95 0.90 0.6

Pseudo-Reduced Volume, τr ??=

1.05

0.70 1.00

0.60

0.50

0.80

5

Nelson-Obert Generalized Compressibility Charts Reduced Pressure, Pr = P Por τ Reduced Temperature, τr = τor

1.10

0.90

τr = 2.5, z = 1.00 τr = 1.5, τr = 3.00 Deviation> 1.0x

0.60

1.20 1.15

1.20

0.9

0.70

3.00 2.00

75 0. 0.10

5 0.6 0.60

0.90

8.00

0.80 0.75 0.10

0.05 Pr

0.7

0.8

0.9

1.0

FIGURE A.1 Compressibility chart for pr < 1.0. (From L. C. Nelson and E. F. Obert, Compressibility Charts. ASME Trans., 76, 1057 ff, 1954.)

Appendix A

931

3.50

1.10 τr,3.00

1.0 0.9 0 0

1.70

0.8 0

2.0 1.3 0 0 1.4 0 1.2 0

1.80

1.60 1.50

0.3

0.70

5

0.4

0

5 0.4

0.5

0

0.80

0.6 0

0.7 0

Compressibility Factor, a = Pv RT

0.90

1.90

0 3.0

5.0

0

2.00

1.00

0.60

0 0.3 1.40 5 0.2

0

0.2

1.30

0.50

Nelson-Obert Generalized Compressibility Charts

1.20

Reduced Pressure, Pr = P Por τ Reduced Temperature, τr = τor

1.15

0.40 1.10

v Pseudo-Reduced Volume, τr ??= Rτo/Por

0.30 0.20 0.0

???

τr,1.00

0.5

1.0

Chart No. 2

1.5

2.0

2.5

3.0 3.5 4.0 Reduced Pressure, Pr

4.5

5.0

5.5

6.0

6.5

7.0

FIGURE A.2 Compressibility chart for 0.50 < pr < 7.0. (From L. C. Nelson and E. F. Obert, Compressibility Charts. ASME Trans., 76, 1057 ff, 1954.)

932

Appendix A

5

0.10

4.0

Nelson-Obert Generalized Compressibility Charts

Compressibility Factor, a = Pv RT

3.0

Reduced Pressure, Pr = P Por τ Reduced Temperature, τr = τor v Pseudo-Reduced Volume, τr ??= Rτo/Por

τr

Chart No. 3

.00 = 1 .05 1 0 1.1 15 1. 0 1.2 0 1.3 0 1.4 0 1.5 0 1.6

5

0.11

0.12

1.80 2.00

???

2.50 3.00

2.0

1.0

0

0.11

3.50 4.00 5.00 6.00 8.00 10.00 15.00

0.13 0.14

0.15 0.20 0.30 0.40

τr = 5.00

0

1.2 .00 1 = τr

0.0 0

5

10

15

20 25 Reduced Pressure, Pr

30

35

40

FIGURE A.3 Compressibility chart for 0 < pr < 40. (From L. C. Nelson and E. F. Obert, Compressibility Charts. ASME Trans., 76, 1057 ff, 1954.)

4

120

S LE U

90

°C

%

JO

ulb

RE

O

80 %

RA TU

(N Y

%

Air

SA TU

50

%

20

40

5 0.5

60

%

RA TI

O

N

TE

20

70

M

A

Day

50

Tem per

atu re ° C

PE

LP

We tB

ram

TH

25

ilog

)K

IL

60

??

EN

K Per ter Me

PE

ENTHALPY = Δn HUMIDITY RATIO ΔW

25

70

0.8

30

20

30

16

?? 90 ??

14

?? ??

12 80

?? ??

10

40

6

4 60 2 45

35

30

25

20

15

10 10

??

50

50

933

FIGURE A.4

5

0°C

6 0.7 0

18

70

Dry Bulb Temperature, °C

2 0.8

0

0

0.8

ity Humid elative 10% R

40

20

%

5 6

?? 100

8

0.84

10

20

??

30 %

10 20

??

8

15

40

%

15

110 22

50

?? ??

Humidity Ration (w) Grams Moisture Per Kilogram Dry Air

IR A M RA G

??

LO

??

80

ic Cub

??

??

??

KI

??

RY

SENSIBLE HEAT = ΔHS TOTAL HEAT ΔHT

?? 24

R

??

??

30

88084˙Book

??

90

me olu 2V 0.0

??

28

??

??

?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ??

D

?? ?? ?? ?? ??

88084

?? ??

13:51

??

0.2

30

Appendix A

28

100

P1: Gopal Joshi

March 22, 2011

?? 30

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Appendix B Summary of Differential Vector Operations in Three Coordinate Systems

y θ

r r

y z

y z

θ

x

x x

z

z (a)

(b)

z

r φ

θ

θ r

y φ x (c) FIGURE B.1 Coordinate systems: (a) Cartesian, (b) cylindrical, and (c) spherical.

B.1

Cartesian Coordinates

Gradient ∇P =

∂P ∂x

x+

∂P ∂y

y+

∂P ∂z

z 935

936

Appendix B

Divergence ∂v x

∇ ·v= Curl

+

∂x

∂v y ∂y

+

∂vz ∂z

⎧  ⎫ ∂vz − ∂vy x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂z ⎪ ⎨  ∂y ⎬  ⎪ ∂vx − ∂vz y ∇ ×v= ∂z ∂x ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ ⎪ ∂ v ∂ y ⎩ − vx z ⎭ ∂x

∂y

Laplacian of a scalar ∇2T =

B.2

∂2 T ∂x2

+

∂2 T ∂ y2

+

∂2 T ∂z2

Cylindrical Coordinates

Gradient ∂P

∇P =

∂r

r+

1 ∂P ∂P + z r ∂ ∂z

Divergence ∇ ·v= Curl

1 ∂ 1 ∂v ∂vz (r vr ) + + r ∂r r ∂ ∂z

⎧ ⎫ 1 ∂vz ∂v ⎪ ⎪ ⎪ ⎪ − r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ r ∂  ∂ z ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨  ∂v ⎬ ∂ v r z ∇ ×v= −  ⎪ ⎪ ∂z ∂r ⎪ ⎪ ⎪ ⎪ ⎪

  ⎪ ⎪ ⎪ ⎪ 1 ∂ ⎪ ∂ v ⎪ ⎪ r ⎪ (r v ) − z⎪ ⎩ ⎭ r ∂r ∂

Laplacian of a scalar 1 ∂ ∇ T= r ∂r 2

B.3

 ∂T 1 ∂2 T ∂2 T r + 2 2 + 2 ∂r r ∂ ∂z

Spherical Coordinates

Gradient ∇P =

∂P ∂r

r+

1 ∂P 1 ∂P +  r ∂ r sin  ∂

Appendix B

937

Divergence ∇ ·v= Curl

1 ∂ 2 1 ∂ 1 ∂v (r vr ) + (v sin ) + r 2 ∂r r sin  ∂ r sin  ∂

⎧   ⎫ 1 ∂ ∂v ⎪ ⎪ ⎪ (v sin ) − r⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ r sin  ∂  ∂  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎨  1 ∂v ⎬ 1 ∂ r ∇ ×v= − (r v )  ⎪ ⎪ r sin  ∂ r ∂r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪1 ∂ ⎪ ∂ v ⎪ ⎪ r ⎪ ⎪ (r v ) −  ⎩ ⎭ r ∂r ∂

Laplacian of a scalar 1 ∂ ∇ T= 2 r ∂r 2

  1 ∂ ∂T 1 ∂2 T 2 ∂T r + 2 sin  + ∂r r sin  ∂ ∂ r 2 sin2  ∂2

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References and Additional Readings

References Achenbach, E. (1968). Distribution of Local Pressure and Skin Friction Around a Circular Cylinder in Cross-Flow up to Re = 5 × 105 . J. Fluid Mech., 34, Part 4. Adamson, A. W. (1960). Physical Chemistry of Surfaces. Interscience, New York. American Society of Heating, Refrigerating and Air-Conditioning Engineers. (1981). ASHRAE Handbook of Fundamentals. ASHRAE, Atlanta. American Society of Heating, Refrigerating and Air-Conditioning Engineers. (1994). Handbook of Refrigeration. ASHRAE, Atlanta. American Society of Mechanical Engineers. (1967). Steam Tables. ASME, New York. American Society of Mechanical Engineers. (1976). ASME Orientation and Guide for Use of SI Units, Guide no. SI-1, 7th ed. ASME, New York. Bar-Cohen, A., and Rohsenow, W. M. (1984). Thermally Optimum Spacing of Vertical Natural Convection Cooled Parallel Plates. J. Heat Trans., 106, 116–122. Bejan, A., and Kraus, A. D. (2003). Handbook of Heat Transfer. John Wiley & Sons, New York, Chap. 4. Blasius, H. (1908). Grenzschichten in Flussigkeiten ¨ mit kleiner Reibung, Z. Math. Phys. Sci., 1. Boelter, L. M. K., Cherry, H., Johnson, H. A., and Martinelli, R. C. (1965). Heat Transfer Notes. McGrawHill, New York. Bowman, R. A., Mueller, A. C., and Nagle, W. M. (1940). Mean Temperature Difference in Design. Trans. ASME, 62, 283–294. Buckingham, E. (1914). On Physically Similar Systems: Illustrations of the Use of Dimensional Equations. Phys. Rev., 4(4), 345–376. Catton, I. (1978). Natural Convection in Enclosures. Proc. Int. Heat Trans. Conf., Toronto, Canada, 6, 13–31. Chenoweth, J. M. (1990). Final Report of the HTRI/TEMA Joint Committee to Review the Fouling Section of the TEMA Standards. Heat Trans. Eng., 11(1), 73–107. Churchill, S. W. (1983). Free Convection Around Immersed Bodies in Heat Exchanger Design Handbook, E. Schlunder, ed., Section 2.5.7. Hemisphere, New York. Churchill, S. W., and Bernstein, M. (1977). A Correlating Equation for Forced Convection from Gases and Liquids to a Circular Cylinder in Cross Flow. J. Heat Trans., 99(2), 300–306. Churchill, S. W., and Chu, H. H. S. (1975a). Correlating Equations for Laminar and Turbulent Free Convection from a Horizontal Cylinder. Int. J. Heat Mass Trans., 18, 1049–1054. Churchill, S. W., and Chu, H. H. S. (1975b). Correlating Equations for Laminar and Turbulent Free Convection from a Vertical Plate. Int. J. Heat Mass Trans., 18, 1323–1328. Churchill, S. W., and Ozoe, H. (1973). Correlations for Laminar Forced Convection with Uniform Heating in Flow Over a Flat Plate and in Developing and Fully Developing Flow in a Tube. J. Heat Trans., 95(1), 78–84. Churchill, S. W., and Usagi, R. (1972). A General Expression for the Correlation of Rates of Heat Transfer and Other Phenomena. AIChe J., 18(6), 1121–1138. Colebrook, C. F. (1939). Turbulent Flow in Pipes with Particular Reference to the Transition Between the Smooth and Rough Pipe Laws, J. Inst. Civil Eng., 11, 133–136. Comstock, J. P., ed. (1967). Principles of Naval Architecture. Society of Naval Architects and Marine Engineers, New York.

939

940

References and Additional Readings

Crane. (1957). Flow of Fluids Through Valves, Fittings and Pipe. Crane Company Technical Paper 410, Chicago. Dittus, F. W., and Boelter, L. M. K. (1930). Univ. Calif. (Berkeley) Publ. Eng., 2, 443. Elenbaas, W. (1942). Heat Distribution of Parallel Plates by Free Convection. Physica, 9(11), 665–671. Fischer, F. K. (1938). Mean Temperature Difference Correction in Multipass Exchangers. Ind. Eng. Chem., 30(4), 377–383. Fourier, J. B. J. (1822). The Analytic Theory of Heat, translated by A. Freeman. Cambridge University Press, Cambridge. Gardner, K. A. (1945). Efficiency of Extended Surfaces. Trans. ASME, 67, 621–628. Gebhart, B., Jaluria, Y., Mahajan, Y. L., and Sammakia, B. (1988). Buoyancy Induced Flows and Transport. Hemisphere, Washington, DC. Goldstein, R. J., Sparrow, E. M., and Jones, D. C. (1973). Natural Convection Mass Transfer Adjacent to Horizontal Plates. Int. J. Heat Mass Trans., 16, 1025–1032. Groeber, ¨ H. (1925). The Heating and Cooling of Simple Geometric Bodies. Z. Ver. Dtsch. Ing., 69, 705–711. Gurney, H. P., and Lurie, J. (1923). Charts for Estimating Temperature Distributions in Heating and Cooling Solid Shapes. Ind. Eng. Chem., 15, 1170–1178. Hausen, H. (1943). Darstellung des W¨armeauberganges in Rohren durch verallgemeinerte Potenzbeziehungen. Z. Ver. Dtsch. Ing., 4, 91. Heisler, M. P. (1947). Temperature Charts for Induction and Constant Temperature Heating. Trans. ASME, 69, 227. Hilpert, R. (1933). W¨armeabgabe von Geheizen Dr¨ahten und Rohren. Forsch. Geb. Ing., 4, 215–224. Hollands, K. G. T. , Unny, S. E., Raithby, G. D., and Konicek, L. (1976). Free Convective Heat Transfer Across Inclined Air Layers. J. Heat Trans., 98, 189–196. Hottel, H. (1954). ”Radiant Heat Transmission,” in W. H. McAdams, ed., Heat Transmission, 3rd ed. McGraw-Hill, New York. Howarth, L. (1938). On the Solution of the Laminar Boundary Layer Equations. Proc. R. Soc. London, AI64. Howell, J. R. (1982). A Catalog of Radiation Configuration Factors. McGraw-Hill, New York. Kakac, S. Shah, R. K., and Aung, W. (1987). Handbook of Single Phase Heat Transfer. John Wiley & Sons, New York. Kays, W. M., and London, A. L. (1984). Compact Heat Exchangers, 3rd ed. McGraw-Hill, New York. Keenan, J. H., Keyes, F. G., Hill, P. G., and Moore, J. G. (1969). Steam Tables. John Wiley & Sons, New York. Kraus, A. D., Aziz, A., and Welty, J. R. (2001). Extended Surface Heat Transfer. John Wiley & Sons, New York. Lloyd, J. R., and Moran, W. R. (1974). Natural Convection Adjacent to Horizontal Surfaces. ASME Paper 74WA/HT-66. American Society of Mechanical Engineers, New York. MacGregor, R. K., and Emery, A. P. (1969), Free Convection Through Vertical Plane Layers: Moderate and High Temperature Fluids. J. Heat Trans., 91, 391–396. Mayinger, F. (1988). Classification and Applications of Two-Phase Flow Heat Exchangers, in TwoPhase Flow Heat Exchangers, S. Kakac, A. E. Bergles, and E. O. Fernandes, eds. Kluwer Academic, Dordrecht, The Netherlands. McAdams, W. H. (1954). Heat Transmission, 3rd ed. McGraw-Hill, New York. McQuiston, F. C., and Tree, D. R. (1972). Optimum Space Envelopes of the Finned Tube Heat Transfer Surface. Trans. ASHRAE, 78, 144–148. Moody, L. F. (1944). Friction Factors for Pipe Flow. Trans. ASME, 66, 681–684. Munson, B. R., Young, D. F, and Okishi, T. K. (1994). Fundamentals of Fluid Mechanics, 2nd ed. John Wiley & Sons, New York. Murray, W. M. (1938). Heat Transfer Through an Annular Disk or Fin of Uniform Thickness. Trans. ASME, J. Appl. Mech., 60, A78–A81. Nagle, W. M. (1933). Mean Temperature Difference in Multipass Exchangers. Ind. Eng. Chem., 25, 604–609. Oppenheim, A. K. (1956). Radiation Analysis by the Network Method. Trans. ASME, 78, 725–732. Planck, M. (1959). The Theory of Heat Radiation. Dover, New York.

References and Additional Readings

941

Raithby, G. D., and Hollands, K. G. T. (1975). A General Method of Obtaining Approximate Solutions to Laminar and Turbulent Free Convection Problems, in Advances in Heat Transfer, T. F. Irvine and J. P. Hartnett, eds. Academic Press, New York, pp. 265–315. Schmidt, E., and Beckmann, W. (1930). Das Temperatur vor einer w¨armeab-gebenden senkrechten Platte bei naturlicher ¨ Convection. Tech. Mech. Thermodyn., 1(10), 341–349; 1(11), 391–160. Schneider, P. J. (1965). Temperature Response Charts. John Wiley & Sons, New York. Sieder, E. N., and Tate, G. E. (1936). Heat Transfer and Pressure Drop of Liquids in Tubes. Ind. Eng. Chem., 28, 1429. Shah, R. K. (1981). Classification of Heat Exchangers, in Heat Exchangers, Thermal–Hydraulic Fundamentals and Design, S. Kakac, A. E. Bergles, and F. Mayinger, eds. Hemisphere, New York. Streeter, V. L. ed. (1961). Handbook of Fluid Dynamics. McGraw-Hill, New York. Underwood, A. J. V. (1934). The Calculation of Mean Temperature Difference in Multipass Heat Exchangers. J. Inst. Petrol. Technol., 20, 145–158. U.S. Standard Atmosphere (1962). Washington, DC: Government Printing Office. U.S. Standard Atmosphere (1976). Washington, DC: Government Printing Office. Whitaker, S. (1972). Forced Convection Heat Transfer Correlations for Flow in Pipes, Past Flat Plates, Single Cylinders, Single Spheres and Flow in Packed Beds and Tube Bundles. AIChE J., 18, 361–371. Zhukauskas, A. (1972). Heat Transfer from Tubes in Cross Flow, in Advances in Heat Transfer, vol. 8, J. P. Hartnett and T. F. Irvine, eds. Academic Press, New York.

Additional Readings Anderson, D. A., Tannehill, J. C., and Pletcher, R. H. (1984). Computational Fluid Mechanics and Heat Transfer. McGraw-Hill, New York. Batchelor, G. K. (1967). An Introduction to Fluid Mechanics. Cambridge University Press, London. Becker, H. A. (1976). Dimensionless Parameters. Halstead Press (Wiley), New York. Bejan, A. (1993). Heat Transfer. John Wiley & Sons, New York. Benedict, R. P. (1984). Fundamentals of Temperature, Pressure, and Flow Measurement. John Wiley & Sons, New York. Black, W. Z., and Hartley, J. G. (1996). Thermodynamics, 3rd ed. Prentice-Hall, Upper Saddle River, NJ. Blevens, R. D. (1984). Applied Fluid Mechanics Handbook. Van Nostrand Reinhold, New York. Bridgeman, P. W. (1922). Dimensional Analysis. Yale University Press, New Haven, CT. Cengel, Y. A. (2003). Heat Transfer: A Practical Approach. 2nd ed. McGraw-Hill, New York. Cengel, Y. A., and Boles, M. A. (1989). Thermodynamics: An Engineering Approach. McGraw-Hill, New York. Currie, I. G. (1974). Fundamental Mechanics of Fluids. McGraw-Hill, New York. Duncan, W. J. (1953). Physical Similarity and Dimensional Analysis; An Elementary Treatise. Edward Arnold, London. Fox, R. W., and McDonald, A. C. (1999). Introduction to Fluid Mechanics, 5th ed. John Wiley & Sons, New York. Gilmer, T. C. (1970). Modern Ship Design. 5, United States Naval Institute, Annapolis, MD, Chap. 5. Goldstein, S. (1938). Modern Developments in Fluid Mechanics. Oxford University Press, London. Happel, J. (1965). Low Reynolds Number Hydrodynamics. Prentice-Hall, Englewood Cliffs, NJ. Hoerner, S. F. (1965). Fluid-Dynamic Drag. Published by the author. Library of Congress No. 64-1966. Holman, J. P. (2002). Heat Transfer, 9th ed. McGraw-Hill, New York. Howell, J. R., and Buckius, R. O. (1992). Fundamentals of Engineering Thermodynamics, 2nd ed. McGrawHill, New York. Huntley, H. E. (1952). Dimensional Analysis, Macdonald, London. Incropera, F. P., and DeWitt, D. P. (2003). Introduction to Heat Transfer, 5th ed. John Wiley & Sons, New York.

942

References and Additional Readings

Ipsen, D. C. (1960). Units, Dimensions and Dimensionless Numbers. McGraw-Hill, New York. Isaacson, E. de St. Q., and Isaacson, M. de St. Q. (1975). Dimensional Methods in Engineering and Physics. John Wiley & Sons, New York. Jepson, R. W. (1976). Analysis of Flow in Pipe Networks. Ann Arbor Science Publishers, Ann Arbor, MI. Kays, W. M., and Crawford, M. E. (1980). Convective Heat and Mass Transfer. McGraw-Hill, New York. Kline, S. J. (1965). Similitude and Approximation Theory. McGraw-Hill, New York. Kraus, A. D. (2001). Matrices for Engineers. Oxford University Press, New York. Kreider, J. F. (1985). Principles of Fluid Mechanics. Allyn & Bacon, Boston. Kuethe, A. M., and Chow, C. Y. (1986). Foundations of Aerodynamics, Bases of Aerodynamics Design, 4th ed. John Wiley & Sons, New York. Langhaar, H. L. (1951). Dimensional Analysis and the Theory of Models. John Wiley & Sons, New York. Mahan, J. R. (2002). Radiation Heat Transfer. John Wiley & Sons, New York. Mills, A. F. (1999). Heat Transfer, 2nd ed. Prentice-Hall, Upper Saddle River, NJ. Milne-Thomson, L. M. (1968). Theoretical Hydrodynamics. Macmillan, New York. Modest, M. (1993). Radiative Heat Transfer. McGraw-Hill, New York. Moran, J. (1984). An Introduction to Theoretical and Computational Aerodynamics. John Wiley & Sons, New York. Moran, M. J., and Shapiro, H. N. (2004). Fundamentals of Engineering Thermodynamics, 5th ed. John Wiley & Sons, New York. Munson, B. R., Young, D. F., and Okiishi, T. (1998). Fundamentals of Fluid Mechanics, 3rd ed. John Wiley & Sons, New York. Murphy, G. (1950). Similitude in Engineering. Ronald Press, New York. Ozisik, M. N. (1985). Heat Transfer: A Basic Approach, McGraw-Hill, New York. Panton, R. L. (1984). Incompressible Flow. Wiley-Interscience, New York. Potter, C., and Wiggert, D. C. (1997). Mechanics of Fluids, 2nd ed. Prentice-Hall, Upper Saddle River, NJ. Reiner, M. (1969). Deformation, Strain and Flow: An Elementary Introduction to Rheology, 3rd ed. Lewis, London. Roberson, J. A., and Grove, C. L. (1997). Engineering Fluid Mechanics. John Wiley & Sons, New York. Rosenhead, L. (1963). Laminar Boundary Layers. Oxford University Press, London. Schlicting, H. (1960). Boundary Layer Theory. McGraw-Hill, New York. Schuring, D. J. (1977). Scale Models in Engineering. Pergamon Press, New York. Sedov, L. I. (1959). Similarity and Dimensional Methods in Mechanics. Academic Press, New York. Siegel, R. and Howell, J. R. (2001). Thermal Radiation Heat Transfer, 4th ed. Taylor & Francis, New York. Sharp, J. J. (1981). Hydraulic Modeling. Butterworth, London. Somerscales, E. F. C., and Knudsen, J. G. (1981). Fouling of Heat Transfer Equipment, Hemisphere, New York. Sonntag, R. E., Borgnakke, C., and VanWylen, G. J. (2003). Fundamentals of Thermodynamics, 6th ed. John Wiley & Sons, New York. Sovran, G., ed. (1978). Aerodynamic Drag Mechanisms of Bluff Bodies and Road Vehicles. Plenum Press, New York. Taylor, E. S. (1974). Dimensional Analysis for Engineers. Clarendon Press, Oxford. Wark, K. Jr., and Richards, D. E. (1999). Thermodynamics, 6th ed. McGraw-Hill, New York. Welty, J. R., Wicks, C. E., Wilson, R. E., and Rorrer, G. L. (2008). Fundamentals of Momentum Heat and Mass Transfer, 5th ed. John Wiley & Sons, New York. White, F. (1974). Viscous Fluid Flow. McGraw-Hill, New York. White, F. M. (1994). Fluid Mechanics, 3rd ed. McGraw-Hill, New York. White, F. M. (1999). Fluid Mechanics, 4th ed. McGraw-Hill, New York. Yalin, M. S. (1971). Theory of Hydraulic Models. Macmillan, London.

Nomenclature

Roman Letter Symbols A AFR AFR a

B Bi b

C

CH CC C∗ CD Cf c

cp cv Cn D

area, m2 air fuel ratio, kg air/kg fuel molal air fuel ratio, kgmol air/kgmol fuel acceleration, m/s2 constant in Redlich-Kwong equation of state, bar-(m3 /kgmol)2 -K1/2 constant in van der Waals equation of state, bar-(m3 /kgmol)2 bulk modulus of elasticity, K−1 Biot number, dimensionless constant in Redlich-Kwong equation of state, m3 /kgmol constant in van der Waals equation of state, m3 /kgmol fin height, m a constant, dimensionless capacitance, Farads capacity rate, W/K hot-side capacity rate, W/K cold-side capacity rate, W/K capacity rate ratio, dimensionless drag coefficient, dimensionless coefficient of skin friction, dimensionless speed of sound, m/s speed of light, 2.99 × 108 , m/s specific heat, dimensionless specific heat at constant pressure, J/kg-K specific heat at constant volume, J/kg-K polytropic specific heat, J/kg-K diameter, m diffusion coefficient, m2 /s

d de E E Eb Eg Eλ El Eu e F

Fc Fn Ft Fx Fy Fz Fo ff G

Gλ Gr g

diameter, m equivalent diameter, m energy, J modulus of elasticity, N/m2 radiant energy emissive power, W/m2 blackbody emissive power, W/m2 -m heat generation, W hemispherical monochromatic emissive power, W/m2 -m Elenbaas number, dimensionless Euler number, dimensionless specific energy, J/kg pipe or tube wall roughness, m emission fraction, dimensionless force, N logarithmic mean temperature difference correction factor, dimensionless shape, view or arrangement factor, dimensionless compression force, N normal force, N tension force, N x-component of force, N y-component of force, N z-component of force, N Fourier modulus, dimensionless Fanning friction factor, dimensionless gravimetric fraction, dimensionless total irradiation, W/m2 hemispherical monochromatic irradiation, W/m2 Grashof number, dimensionless acceleration of gravity, m/s2 943

944 proportionality constant, 32.174, lbm -ft/s2 -lbf H enthalpy, J height, m hp horsepower h specific enthalpy, J/kg heat transfer coefficient, W/m2 -K head, m hL head lost, m l current, amperes intensity of radiation, W/m2 -sr moment of inertia, m4 Io moment of inertia about centroidal axis, m4 i instantaneous electric current, amperes J Bessel function of first kind, dimensionless total radiosity, W/m2 Jλ monochromatic radiosity, W/m2 -m KE kinetic energy, J k Boltzmann constant ratio of specific heats, dimensionless thermal conductivity, W/m-K mass transfer coefficient, m/s ke specific kinetic energy, J/kg kg spring constant, kg/m L liters, m3 length, m inductance, H Le entrance length, m LMTD logarithmic mean temperature difference, K M molecular weight, kg/kgmol Mach number, dimensionless moment, N-m m mass, kg fin performance factor, m−1 m ˙ mass flow rate, kg/s Nu Nusselt number, dimensionless Ntu number of transfer units. dimensionless n number of moles, dimensionless number of tubes, dimensionless number of fins, dimensionless rotational speed, rpm P pressure, N/m2 (Pa) power, W fin perimeter, m

Nomenclature

gc

Pcr pr PE Pr p pr pe Q ˙ Q q qi q˙ R

Rd ¯ R Ra Re r

re rc rp S SG s T

Tas Tc T∞ t

U

u V

thermodynamic probability, dimensionless critical pressure, N/m2 (Pa) reduced pressure, dimensionless potential energy, J Prandtl number, dimensionless number of dependent intrinsic variables, dimensionless reference pressure, N/m2 (Pa) specific internal energy, J/kg heat, J heat flow, W instantaneous charge, Coulomb heat generated per unit volume, W/m3 heat flux, W/m2 gas constant, J/kg-K radius of gyration, m radiation resistance, m−2 fouling resistance, K/W universal gas constant, 8314 J/kgmol-K Rayleigh number, dimensionless Reynolds number, dimensionless radius, m radial coordinate, m compression ratio, dimensionless cutoff ratio, dimensionless critical radius, m pressure ratio, dimensionless surface area, m2 entropy, J/K specific gravity, dimensionless specific entropy, J/kg-K Temperature, units vary tensile force, N torque N-m adiabatic saturation temperature, K critical temperature, K free stream or surrounding temperature, K time, sec exchanger cold-side temperature, units vary internal energy, J overall heat transfer coefficient, W/m2 -K specific internal energy, J/kg volume, m3 voltage, volts

Nomenclature Vˆ Vˆ ∞ V˙ v v¯ W ˙ W We Wp Ws

velocity, m/s free-stream velocity, m/s volumetric flow rate, m3 /s specific volume, m3 /kg molal specific volume, m3 /kgmol weight, N work, J width, m power, W electrical work, J paddle wheel work, J spring work, J

945 x

Y y

Z z

length coordinate, m quality, kg vapor/kg mixture, percent volumetric fraction, dimensionless length coordinate, m moisture fraction, kg vapor/kg mixture, percent centroidal axis location, m compressibility factor, dimensionless length coordinate, m intrinsic variable, dimensionless plate spacing, m

Greek Letter Symbols α

β

γ

Δ ΔT δ

δW δQ 

η

angular acceleration, rad/s2 absorbtivity, dimensionless angle, degree or radian thermal diffusivity, m2 /s refrigerator coefficient of performance, dimensionless volumetric coefficient of expansion, K−1 angle, degree or radian aircraft attitude angle, degree or radian heat pump coefficient of performance, dimensionless specific weight, N/m3 change in, dimensionless gap thickness, m time interval, s inexact differential, dimensionless fin thickness, m boundary layer thickness, m static deflection, m differential work, J differential heat, J emissivity, dimensionless heat exchanger effectiveness, dimensionless fin effectiveness, dimensionless efficiency, dimensionless fin efficiency, dimensionless

θ

θm Λ λ μ v ρ ρe σ

τ

γ φ ψ Ω ω

temperature difference, K angular displacement, radians wetting angle, degrees or radians mean temperature difference, K aspect ratio, dimensionless wavelength, m eigenvalue, dimensionless dynamic viscosity, kg/m-s kinematic viscosity, m2 /s density, kg/m3 reflectivity, dimensionless electrical resistivity, -m Stefan-Boltzmann constant, 5.67 × 10−8 , W/m2 -K4 normal stress component, N/m2 surface tension, N/m torque, N-m shear stress, N/m2 tangential shear component, N/m2 transmissivity, dimensionless time constant, s percent throretical air an angle, degrees or radians wet bulb depression, K electrical resistance, Ohm angular velocity, rad/s specific humidity, kg water vapor/kg dry air solid angle, steradians

946

Nomenclature

Roman Subscripts avg atm C

c cv cyc da dirt eng e flow f fg

gage gen g H i id

indicates average condition indicates atmospheric pressure indicates Carnot indicates cold or cold side indicates compressor indicates critical condition indicates compression indicates control volume indicates cycle indicates dry air indicates dirt or fouling indicates engine indicates exit or exiting condition indicates equivalent indicates flow indicates saturated liquid condition indicates difference between saturated liquid and saturated vapor indicates gage pressure indicates generator or generated value indicates saturated vapor indicates hot or hot side indicates in or inlet condition indicates inner or inside indicates ideal

in indicates input irrev indicates irreversible liq indicates liquid L indicates low or low side max indicates a maximum min indicates a minimum mix indicates mixture nonflow indicates nonflow condition n indicates normal component o indicates original or nominal value indicates outer or outside opt indicates optimum condition out indicates output proj indicates projected P indicates pump ref indicates reference condition rev indicates reversible r indicates reduced condition sat indicates saturated condition sink indicates sink source indicates source t indicates tension total indicates total trans indicates translational v indicates vapor vap indicates vapor vac indicates vacuum wv indicates water vapor

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